Hello aspiring mathematicians! Welcome to the exciting world of Coordinate Geometry. Today, we're going to unravel a concept that is absolutely fundamental to understanding shapes and curves in mathematics: Locus and its equation. Don't let the fancy word "locus" intimidate you; it's a very intuitive idea once we break it down.

### What is Locus? The Path of a Point!

Imagine you're walking. Every step you take creates a path. If you follow certain rules while walking, your path will take on a specific shape. For example, if you walk in a way that you're always exactly 5 meters away from a specific tree, what kind of path would you trace? A circle, right?

That's essentially what "locus" is! In mathematics, a locus is simply the path traced by a point (or set of points) that moves according to one or more specific geometric conditions.

Think of it like this:

* You are the "point" P(x, y).
* Your movement rules are the "geometric conditions."
* The path you trace is the "locus."

So, whenever you hear "locus," just think of it as a "collection of all points that satisfy a given condition."

Let's use a couple of simple analogies to build our intuition:

#### Analogy 1: The Ant on a String

Imagine you tie a very short string to a pole in the ground, and tie the other end to an ant. If the ant walks around, keeping the string taut (stretched), what kind of path will it trace?







It will trace a perfect circle! Here:
* The ant is our moving point P(x, y).
* The pole is a fixed point.
* The string length is the fixed distance.
* The condition is: "the ant is always at a fixed distance from the pole."
* The locus is the circle itself – the path traced by the ant.

#### Analogy 2: The Airplane's Flight Path

When an airplane flies, its flight path can be tracked. If air traffic control asks a pilot to always maintain a constant altitude (height) of, say, 10,000 meters above the ground, what kind of surface would the plane fly on? It would be a flat plane parallel to the ground! Here:
* The plane is our moving point.
* The ground is a fixed reference.
* The condition is: "the plane is always at a constant distance from the ground."
* The locus is a plane parallel to the ground.

### The Two Golden Rules of Locus

For a collection of points to be considered the locus for a given condition, it must satisfy two crucial requirements:

1. Every point that satisfies the given condition MUST lie on the locus.
2. Every point that lies on the locus MUST satisfy the given condition.

If either of these rules is broken, then you haven't correctly identified the locus!

### Common Examples of Locus (Geometric Interpretation)

Let's look at some basic geometric shapes and understand them as a locus:

1. Locus of points equidistant from a fixed point:
* Fixed Point: Let's call it O.
* Condition: Any point P must always be the same distance (say, 'r') from O.
* Locus: A circle with O as its center and 'r' as its radius.

2. Locus of points equidistant from two fixed points:
* Fixed Points: Let's call them A and B.
* Condition: Any point P must always be such that its distance from A is equal to its distance from B (PA = PB).
* Locus: The perpendicular bisector of the line segment AB. Remember, any point on the perpendicular bisector is equidistant from the endpoints of the segment!

3. Locus of points equidistant from a fixed line:
* Fixed Line: Let's call it 'L'.
* Condition: Any point P must always be the same distance (say, 'd') from line L.
* Locus: A pair of parallel lines, one on each side of L, both at a distance 'd' from L.

4. Locus of points forming a given angle with a line segment:
* Fixed Line Segment: Let's call it AB.
* Condition: Any point P must always be such that the angle APB is constant (e.g., 90 degrees).
* Locus: If the angle is 90 degrees, the locus is a circle with AB as its diameter (excluding points A and B themselves). If it's another angle, it's a part of a circle.

These examples clearly show that common geometric figures are nothing but the loci of points satisfying specific conditions.

### The Equation of a Locus: Bringing Algebra into the Picture

So far, we've talked about locus geometrically. But in coordinate geometry, we want to describe these paths using algebra. This is where the equation of a locus comes into play.

An equation of a locus is an algebraic expression that is satisfied by the coordinates (x, y) of every single point on the locus, and by no other point.

Think of it as a mathematical "ID card" for the path. If a point (x, y) satisfies this equation, it's on the path. If it doesn't, it's not.

#### How do we find this magical equation?

The process is quite straightforward! We use our knowledge of coordinate geometry tools like the distance formula, section formula, slope formula, etc., to translate the geometric condition into an algebraic equation.

Here are the general steps:

1. Assume a General Point: Let P(x, y) be any general point that lies on the locus. This point P(x, y) is the "ant" or "airplane" we discussed.
2. Write Down the Given Condition: Clearly state the geometric condition(s) that P(x, y) must satisfy.
3. Translate into an Algebraic Equation: Use the coordinates (x, y) of P and the coordinates of any other given fixed points/lines to express the geometric condition algebraically.
* If the condition involves distance, use the distance formula.
* If it involves a midpoint or ratio, use the section formula.
* If it involves perpendicularity or parallelism, use slopes.
* If it involves areas, use the area formula.
4. Simplify the Equation: Rearrange and simplify the algebraic equation to get it into a standard and neat form. This simplified equation is the required equation of the locus.

### Let's try a simple example to see this in action!

Example: Find the equation of the locus of a point P such that its distance from the origin O(0, 0) is always 5 units.

Step-by-step Solution:

1. Assume a General Point: Let P(x, y) be any point on the locus.

Point	Coordinates
P	(x, y)
Origin (O)	(0, 0)

2. Write Down the Given Condition: The problem states that the distance of point P from the origin O is always 5 units.
So, PO = 5.

3. Translate into an Algebraic Equation: We need to use the distance formula between P(x, y) and O(0, 0).
The distance formula is: $d = sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
Applying this, $PO = sqrt{(x - 0)^2 + (y - 0)^2}$
So, $sqrt{x^2 + y^2} = 5$

4. Simplify the Equation: To get rid of the square root, we square both sides of the equation:
$(sqrt{x^2 + y^2})^2 = 5^2$
$x^2 + y^2 = 25$

This is the equation of the locus! Do you recognize it? It's the standard equation of a circle centered at the origin with a radius of 5 units. Every point (x, y) on this circle satisfies the condition, and any point not on this circle will not satisfy it. Perfect!

### Why is Locus so Important? (A Peek Ahead)

Understanding locus is not just an academic exercise. It's the very foundation for studying almost all geometric figures in coordinate geometry.

* A circle is the locus of points equidistant from a fixed point.
* A parabola is the locus of points equidistant from a fixed point (focus) and a fixed line (directrix).
* An ellipse is the locus of points such that the sum of their distances from two fixed points (foci) is constant.
* A hyperbola is the locus of points such that the absolute difference of their distances from two fixed points (foci) is constant.

See? Locus helps us *define* these shapes algebraically, which allows us to analyze their properties, solve problems, and even design things like satellite dishes (parabolas) or planetary orbits (ellipses).

JEE Focus: While the fundamental definition of locus is simple, JEE problems often involve complex conditions. You'll need to be adept at using all coordinate geometry formulas (distance, section, slope, area, perpendicular distance from a point to a line, etc.) to translate these conditions into equations. The simplification step can also become quite involved, requiring strong algebraic manipulation skills.

### Let's Recap!

* A locus is the path or collection of all points that satisfy one or more given geometric conditions.
* Every point on the locus *must* satisfy the condition, and every point satisfying the condition *must* be on the locus.
* The equation of a locus is an algebraic equation that describes this path.
* To find the equation, we:
1. Assume P(x, y) is a general point on the locus.
2. Write down the given condition(s).
3. Translate these conditions into an algebraic equation using relevant coordinate geometry formulas.
4. Simplify the equation.

This fundamental understanding will be your stepping stone to mastering many advanced concepts in coordinate geometry. Keep practicing, and you'll soon be tracing complex paths with ease!

Welcome back, future engineers! In our previous discussions, we established the fundamental idea of a Locus – a path traced by a moving point under certain specified geometric conditions. Now, let's roll up our sleeves and dive much deeper into the mechanics of finding the equation of a locus, understanding the nuances, and tackling problems that often appear in JEE Mains and Advanced.

1. Refining the Definition of Locus

In a more formal mathematical sense, a locus is defined as the set of all points, and only those points, which satisfy a given geometric condition or set of conditions. The "equation of the locus" is the algebraic expression that represents this set of points in a coordinate system. It's a precise mathematical statement that every point on the locus satisfies, and no point not on the locus satisfies.

• "All points": This means the equation must cover every single point that meets the criteria.


• "Only those points": This means the equation should not include any point that *doesn't* meet the criteria.

2. The Universal Procedure for Finding the Equation of a Locus

Regardless of the complexity, the core strategy remains consistent. Let's outline it step-by-step, paying close attention to the algebraic translation.

1. Assume the Moving Point: Let the coordinates of the moving point, whose locus is to be found, be P(h, k). We use (h, k) instead of (x, y) initially to differentiate it from other fixed or variable points in the problem.


2. Translate Geometric Condition(s) into Algebraic Equations: Express the given geometric condition(s) involving P(h, k) and any other fixed points, lines, or variables in algebraic form using the distance formula, section formula, slope formula, etc.


3. Simplify and Eliminate Parameters: Simplify the algebraic equation(s). If any intermediate variables or parameters were introduced (e.g., a variable angle, a point on another curve), eliminate them to obtain an equation solely in terms of h and k. This is a critical step, especially for JEE Advanced problems.


4. Replace (h, k) with (x, y): Finally, replace 'h' with 'x' and 'k' with 'y' to get the equation of the locus in standard coordinate form.

This procedure is robust. Let's see it in action with some fundamental and advanced examples.

3. Deeper Look at Common Loci Derivations

Example 1: Locus of a point equidistant from two fixed points.

Let the two fixed points be A(x₁, y₁) and B(x₂, y₂). Let the moving point be P(h, k).

1. Condition: P is equidistant from A and B. So, PA = PB.


2. Algebraic Translation:
   
   Using the distance formula, $PA^2 = PB^2$ (squaring simplifies calculations).
   
   $(h - x₁)^2 + (k - y₁)^2 = (h - x₂)^2 + (k - y₂)^2$
   


3. Simplification:
   
   $h^2 - 2hx₁ + x₁^2 + k^2 - 2ky₁ + y₁^2 = h^2 - 2hx₂ + x₂^2 + k^2 - 2ky₂ + y₂^2$
   
   Notice $h^2$ and $k^2$ terms cancel out, which is a common characteristic for linear loci.
   
   $-2hx₁ - 2ky₁ + x₁^2 + y₁^2 = -2hx₂ - 2ky₂ + x₂^2 + y₂^2$
   
   Rearrange terms to group h and k:
   
   $2h(x₂ - x₁) + 2k(y₂ - y₁) + (x₁^2 + y₁^2 - x₂^2 - y₂^2) = 0$
   


4. Replace (h, k) with (x, y):
   
   $2x(x₂ - x₁) + 2y(y₂ - y₁) + (x₁^2 + y₁^2 - x₂^2 - y₂^2) = 0$
   

This is the equation of a straight line. Geometrically, this line is the perpendicular bisector of the line segment AB. Every point on this line is equidistant from A and B, and every point equidistant from A and B lies on this line.

Example 2: Locus of a point equidistant from a fixed point (Focus) and a fixed line (Directrix).

Let the fixed point be F(a, 0) (for simplicity, often taken as the focus).
Let the fixed line be x = -a (the directrix, for simplicity).
Let the moving point be P(h, k).

1. Condition: Distance from P to F equals distance from P to the line x = -a. Let M be the foot of the perpendicular from P to the line x = -a. Then PF = PM.


2. Algebraic Translation:
   
   $PF^2 = PM^2$
   
   The distance from P(h, k) to F(a, 0) is $sqrt{(h - a)^2 + (k - 0)^2}$.
   
   The distance from P(h, k) to the line x + a = 0 is $frac{|h + a|}{sqrt{1^2 + 0^2}} = |h + a|$.
   
   So, $(h - a)^2 + k^2 = (h + a)^2$
   


3. Simplification:
   
   $h^2 - 2ah + a^2 + k^2 = h^2 + 2ah + a^2$
   
   Cancel $h^2$ and $a^2$ terms:
   
   $-2ah + k^2 = 2ah$
   
   $k^2 = 4ah$
   


4. Replace (h, k) with (x, y):
   
   $y^2 = 4ax$
   

This is the standard equation of a parabola. This derivation is fundamental to understanding parabolic curves.

Example 3: Locus of a point such that the ratio of its distances from two fixed points is a constant. (Apollonius' Circle)

Let the two fixed points be A(x₁, y₁) and B(x₂, y₂). Let the moving point be P(h, k).
Let the constant ratio be $lambda$ (where $lambda > 0$).

1. Condition: $frac{PA}{PB} = lambda implies PA = lambda PB$.


2. Algebraic Translation:
   
   $PA^2 = lambda^2 PB^2$
   
   $(h - x₁)^2 + (k - y₁)^2 = lambda^2 [(h - x₂)^2 + (k - y₂)^2]$
   


3. Simplification:
   
   $h^2 - 2hx₁ + x₁^2 + k^2 - 2ky₁ + y₁^2 = lambda^2 [h^2 - 2hx₂ + x₂^2 + k^2 - 2ky₂ + y₂^2]$
   
   $(1 - lambda^2)h^2 + (1 - lambda^2)k^2 - 2h(x₁ - lambda^2 x₂) - 2k(y₁ - lambda^2 y₂) + (x₁^2 + y₁^2 - lambda^2(x₂^2 + y₂^2)) = 0$
   


4. Replace (h, k) with (x, y):
   
   $(1 - lambda^2)x^2 + (1 - lambda^2)y^2 - 2x(x₁ - lambda^2 x₂) - 2y(y₁ - lambda^2 y₂) + (x₁^2 + y₁^2 - lambda^2(x₂^2 + y₂^2)) = 0$
   

This is the general equation of a circle, provided $lambda
eq 1$.

• If $lambda = 1$: The equation simplifies to $0 cdot x^2 + 0 cdot y^2 - 2x(x₁ - x₂) - 2y(y₁ - y₂) + (x₁^2 + y₁^2 - x₂^2 - y₂^2) = 0$. This is the same linear equation we derived for the perpendicular bisector in Example 1, which is a straight line.


• If $lambda
  eq 1$: The coefficients of $x^2$ and $y^2$ are equal and non-zero, indicating a circle. This is known as Apollonius' Circle.

4. JEE-Specific Considerations: Parametric Locus and Elimination

Many challenging locus problems in JEE involve a moving point whose coordinates are expressed in terms of a parameter (like 't' or an angle 'θ'). In such cases, the process involves an additional crucial step: eliminating the parameter.

Example 4: Locus of the centroid of a triangle whose vertices are A(a cosθ, a sinθ), B(b, 0), and C(c, 0).

Here, the first vertex A is itself moving on a circle $x^2 + y^2 = a^2$ as $ heta$ varies. B and C are fixed points.

1. Assume the Moving Point: Let the centroid be G(h, k).


2. Translate Geometric Condition: The centroid G of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by $G = left(frac{x_1+x_2+x_3}{3}, frac{y_1+y_2+y_3}{3}
   ight)$.
   
   So, for our triangle:
   
   $h = frac{a cos heta + b + c}{3}$
   
   $k = frac{a sin heta + 0 + 0}{3}$
   


3. Simplify and Eliminate Parameters:
   
   From the equations above, we have expressions for $cos heta$ and $sin heta$:
   
   $3h - (b + c) = a cos heta implies cos heta = frac{3h - (b + c)}{a}$
   
   $3k = a sin heta implies sin heta = frac{3k}{a}$
   
   Now, we use the identity $cos^2 heta + sin^2 heta = 1$ to eliminate $ heta$:
   
   $left(frac{3h - (b + c)}{a}
   ight)^2 + left(frac{3k}{a}
   ight)^2 = 1$
   
   $(3h - (b + c))^2 + (3k)^2 = a^2$
   


4. Replace (h, k) with (x, y):
   
   $(3x - (b + c))^2 + (3y)^2 = a^2$
   

This is the equation of a circle. The locus of the centroid is a circle whose center is $left(frac{b+c}{3}, 0
ight)$ and radius is $frac{a}{3}$. This illustrates how parametric elimination is vital for locus problems.

5. Important Nuances and JEE Traps

• Exclusions/Restrictions: Sometimes, the geometric condition might implicitly exclude certain points. For example, if a point P divides a segment AB in a ratio m:n, then P cannot be A or B. Or if it's "distance from P to line L" where P is on L, the distance is 0. Be mindful of degenerate cases or points that violate initial assumptions (e.g., denominators becoming zero).


• Completeness and Correctness: Ensure your derived equation represents *all* points satisfying the condition and *only* those points. Squaring operations, while simplifying, can sometimes introduce extraneous solutions (e.g., $x=2$ leads to $x^2=4$, but $x^2=4$ also means $x=-2$). Always verify the final equation against the original condition.


• Interpreting the Result: After finding the algebraic equation, try to recognize the curve it represents (straight line, circle, parabola, ellipse, hyperbola). This helps in cross-checking your answer and understanding the geometric implications.


• Use of Auxiliary Points/Lines: Sometimes, introducing an auxiliary point (like a midpoint or a foot of perpendicular) or an auxiliary line can simplify the translation of conditions, especially in problems involving angles or specific geometric constructions.

Aspect	CBSE/State Board Focus	JEE Main & Advanced Focus
Complexity of Conditions	Generally straightforward distance or ratio conditions.	Involves multiple conditions, angles, areas, fixed/variable points, and parametric relationships.
Derivations	Focus on standard derivations like circle, parabola (from basic definitions).	Expect to derive locus for more complex scenarios, including those involving parameters that need elimination.
Problem Types	Direct application of locus definition, often leading to standard forms.	Problems requiring clever algebraic manipulation, elimination of parameters, identification of hidden geometric properties, and handling of edge cases.
Key Skill	Algebraic translation of simple geometric conditions.	Strong command over coordinate geometry formulas, algebraic manipulation, trigonometric identities for parameter elimination, and geometric reasoning.

Mastering the concept of locus and its equation is foundational for much of coordinate geometry. It's not just about memorizing formulas, but about developing a systematic approach to translate geometric thoughts into algebraic expressions. Keep practicing, and you'll find that many seemingly complex problems boil down to this very procedure!

Welcome to the 'Mnemonics and Short-cuts' section for Locus and its Equation! While understanding the core concepts is paramount, a few handy memory aids and quick identification strategies can significantly speed up your problem-solving process in exams.

Mnemonic for the Process of Finding Locus Equation

Finding the equation of a locus involves a systematic approach. Remember the acronym "APES" to recall the steps:

• Assume P(x, y) as the moving point.


• Put down the geometric Properties or conditions given in the problem.


• Equate these conditions algebraically to form an Equation.


• Simplify the equation to its standard form.

This "APES" method ensures you follow all necessary steps without missing any, leading you to the correct locus equation efficiently.

Quick Identification Guide for Common Loci

Many locus problems relate to standard geometric shapes. Recognizing these conditions instantly can save time. Here's a table summarizing common conditions and their resulting loci:

Condition	Resulting Locus
Distance from a fixed point is constant.	Circle
Equidistant from two fixed points.	Perpendicular Bisector of the line segment joining the two points.
Equidistant from two fixed lines.	Angle Bisectors of the angles formed by the lines.
Equidistant from a fixed point (focus) and a fixed line (directrix).	Parabola
Sum of distances from two fixed points (foci) is constant.	Ellipse
Difference of distances from two fixed points (foci) is constant.	Hyperbola

JEE Tip: For problems involving distances, immediately recognize the underlying standard form. For instance, if you see PA = PB (distances from two points are equal), you know it will simplify to a linear equation, representing a perpendicular bisector, even before algebraic manipulation.

Shortcut for Distance Calculations

When dealing with distances, especially using the distance formula, you'll often encounter square roots. A simple shortcut for algebraic manipulation is to:

• Always square both sides of the equation immediately if the condition involves equality of distances (e.g., PA = PB, PA = k * PB). This eliminates the square roots early on, simplifying the algebra significantly and reducing the chances of errors.

For example, if the condition is $sqrt{(x-x_1)^2 + (y-y_1)^2} = sqrt{(x-x_2)^2 + (y-y_2)^2}$, square both sides to get $(x-x_1)^2 + (y-y_1)^2 = (x-x_2)^2 + (y-y_2)^2$. This simplifies much faster.

By using these mnemonics and shortcuts, you can approach locus problems with greater confidence and efficiency, ensuring accuracy and saving precious time during your exams. Keep practicing!

Quick Tips for Locus and its Equation

Understanding and quickly solving problems involving locus is fundamental in Coordinate Geometry. This section provides exam-oriented tips to master finding the equation of a locus.

1. Grasp the Definition of Locus

• The locus of a point is the path traced by a point that moves under one or more specific geometric conditions.


• Essentially, it's a set of all points that satisfy a given property.

2. Standard Procedure to Find the Locus Equation

Follow these systematic steps to determine the equation of a locus:

1. Assume the Moving Point: Let the coordinates of the moving point be P(h, k). This point is what traces the locus.


2. Write Down the Geometric Condition: Express the given conditions involving P(h, k) and other fixed points, lines, or quantities. This might involve distances, slopes, areas, or ratios.


3. Translate to Algebraic Form: Convert the geometric condition into an algebraic equation using relevant formulas such as:
   
   
   
   • Distance formula: $sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$
   
   
   • Section formula: $(frac{m x_2 + n x_1}{m+n}, frac{m y_2 + n y_1}{m+n})$
   
   
   • Slope formula: $frac{y_2-y_1}{x_2-x_1}$
   
   
   • Area formula for triangles, or properties of perpendicular/parallel lines.
   
   
   
   


4. Simplify the Equation: Perform necessary algebraic manipulations to simplify the equation. Ensure that 'h' and 'k' are the only variables related to the moving point.


5. Replace (h, k) with (x, y): Finally, replace 'h' with 'x' and 'k' with 'y' to obtain the standard equation of the locus. This represents all points (x, y) that satisfy the given conditions.

3. Identify Key Information in the Problem

Always look for the following clues in the problem statement:

• Fixed points: Coordinates will be given (e.g., A(1, 2), B(3, 4)).


• Fixed lines: Equations will be given (e.g., $x+y=5$).


• Fixed distances or ratios: A constant value (e.g., distance is 5 units, ratio of distances is 2:1).


• Geometric relationships: Perpendicularity, parallelism, collinearity, area constraints.

4. Common Loci and Their Characteristics (JEE & CBSE)

Recognizing these can often short-circuit the derivation process:

• Perpendicular Bisector: Locus of a point equidistant from two fixed points. (Equation is a straight line).


• Circle: Locus of a point whose distance from a fixed point (center) is constant (radius). (Equation is $ (x-a)^2 + (y-b)^2 = r^2 $).


• Parabola: Locus of a point equidistant from a fixed point (focus) and a fixed line (directrix). (Standard equations like $y^2 = 4ax$ or $x^2 = 4ay$).


• Ellipse/Hyperbola: Locus of a point such that the sum/difference of its distances from two fixed points (foci) is constant. (Standard forms $x^2/a^2 + y^2/b^2 = 1$ or $x^2/a^2 - y^2/b^2 = 1$).


• Apollonius Circle: Locus of a point such that the ratio of its distances from two fixed points is a constant (not equal to 1). This is always a circle.

5. JEE vs. CBSE Approach

Aspect	CBSE Board Exams	JEE Main/Advanced
Complexity of Conditions	Generally straightforward, leading to basic lines or circles.	Can be complex, involving multiple conditions, leading to general conics or more intricate curves. Often requires strong algebraic manipulation.
Typical Loci	Lines, Circles.	All conics (parabola, ellipse, hyperbola), straight lines, circles. May also involve parametric forms or properties of tangents/normals.
Focus	Understanding the concept and applying basic formulas correctly.	Efficient application of formulas, advanced algebraic simplification, recognizing standard forms, and integrating with other topics.

6. Quick Checks & Warnings

• Don't forget to replace (h, k) with (x, y) at the end. This is a common oversight.


• Always check if the derived equation makes sense geometrically. For example, if the condition implies a straight line, your final equation should be linear.


• Be meticulous with algebraic squaring, especially when dealing with distance formulas to eliminate square roots. A common error involves signs.


• Ensure that all variables representing fixed points or other parameters are constants, not variables related to the moving point.

Mastering locus problems lays a strong foundation for understanding conic sections and other advanced topics in coordinate geometry. Practice these tips to enhance your problem-solving speed and accuracy!

What is Locus? An Intuitive Approach

Imagine a point moving in a plane, but not just randomly. It moves in a way that always satisfies a specific rule or condition. The path traced by this point, or the set of all possible positions it can occupy while adhering to that rule, is called its locus.

Think of it like this:

• A dog tied to a pole with a rope. The dog can move, but its distance from the pole must always be equal to the length of the rope. The path it traces is a circle. This circle is the locus of the dog's position.


• A car driving on a straight road. All points on the road represent the positions where the car can be. The straight line is the locus of the car's position.

In essence, a locus is the geometric path or curve formed by a point that moves according to a specific geometric condition.

The Core Idea: Condition is King!

The most crucial aspect of understanding locus is the geometric condition. Every single point that belongs to the locus MUST satisfy this condition, and conversely, any point that satisfies the condition MUST belong to the locus. If a point does not satisfy the condition, it cannot be on the locus.

For example, if the condition is "the distance from point A is equal to the distance from point B," then every point on the locus will be equidistant from A and B (this locus is a perpendicular bisector). Any point not on the perpendicular bisector will not be equidistant from A and B.

From Geometric Condition to Algebraic Equation

In coordinate geometry, we represent points using coordinates (x, y). When a point P(x, y) moves satisfying a given geometric condition, we can translate this condition into an algebraic relationship between x and y. This algebraic relationship is called the equation of the locus.

The goal in "finding the locus" is to express the geometric condition in the language of algebra (an equation) in terms of x and y.

The Process in a Nutshell

Finding the equation of a locus generally involves these steps:

1. Assume a Moving Point: Let P(x, y) be the coordinates of the moving point whose locus is to be found.


2. Write the Geometric Condition: Express the given geometric condition mathematically using the coordinates of P(x, y) and any other given points/lines.


3. Simplify to an Equation: Simplify the mathematical expression into an algebraic equation involving x and y. This simplified equation is the equation of the locus.


4. Identify the Shape (Optional but often useful): Once you have the equation, you might recognize it as a specific geometric shape (e.g., a straight line, circle, parabola, ellipse, hyperbola).

Why is this Important? (JEE & CBSE Relevance)

Understanding locus is fundamental to coordinate geometry because many standard geometric shapes (like lines, circles, parabolas, ellipses, and hyperbolas) are essentially loci of points satisfying specific conditions.

Exam Focus	Relevance to Locus
JEE Main Focus	Locus is a foundational concept. Questions often directly ask to find the locus of a point satisfying certain conditions, which can lead to equations of lines, circles, or conic sections. A strong intuitive grasp helps in correctly setting up the problem.
CBSE Board Exam Focus	While direct "find the locus" questions are less common, the underlying principle is crucial for understanding the derivations of standard equations for lines, circles, and conic sections. It helps build a deeper conceptual understanding of these shapes.

Keep your intuition sharp! When you encounter a locus problem, always ask yourself: "What is the *rule* this point must follow?" and then translate that rule into an equation.

Understanding the concept of Locus and its equation isn't just an abstract mathematical exercise; it forms the backbone of countless real-world applications across science, engineering, and technology. By defining the path traced by a point under specific geometric conditions, we can model, predict, and design complex systems.

Here are some key real-world applications of Locus:

• 
  Planetary and Satellite Orbits (Ellipses):
  

  One of the most profound applications is found in astronomy. According to Kepler's First Law, planets orbit the sun in an elliptical path, with the sun at one of the foci. Similarly, man-made satellites, including those for communication or GPS, often follow elliptical or circular orbits around Earth. The equation of an ellipse helps scientists and engineers accurately predict and track these trajectories, ensuring successful missions and reliable communication.

  
  

  JEE Relevance: Questions involving properties of ellipses (foci, eccentricity) are directly linked to understanding such orbital mechanics.

  
  


• 
  Antenna, Headlight, and Solar Reflector Design (Parabolas):
  

  The reflective property of a parabola is critical in various designs. A parabolic dish antenna (like a satellite dish) is shaped such that all incoming parallel signals reflect towards a single point, the focus, where the receiver is placed. Conversely, in car headlights or searchlights, a light source placed at the focus of a parabolic reflector emits a strong, parallel beam of light. Solar concentrators also use parabolic shapes to focus sunlight onto a single point for energy generation.

  
  

  CBSE & JEE Relevance: Understanding the focus, directrix, and reflective properties of parabolas is a fundamental concept tested in both curricula.

  
  


• 
  GPS and Navigation (Circles/Spheres):
  

  Global Positioning Systems (GPS) heavily rely on the concept of locus. When your GPS device receives signals from multiple satellites, each signal provides your distance from that specific satellite. The locus of points equidistant from a single satellite is a sphere. By receiving signals from at least three (or four for 3D positioning) satellites, your device determines your precise location as the intersection point of these spheres. This is an application of "trilateration," which is essentially finding the common point of intersection of loci.

  
  


• 
  Architectural and Structural Design (Various Conics):
  

  Architects and civil engineers frequently use conic sections in their designs. Arch bridges often take the shape of parabolas or ellipses for structural stability and aesthetic appeal. The strength and distribution of forces in such structures are optimized by these geometric forms. For instance, a parabolic arch distributes the weight uniformly, minimizing stress points.

  
  


• 
  Acoustic and Optical Applications (Ellipses):
  

  The reflective property of ellipses is used in whispering galleries, where a whisper at one focus can be heard clearly at the other focus, even across a large room. In medicine, elliptical reflectors are used in lithotripsy machines to focus shock waves onto kidney stones, breaking them without invasive surgery. The patient is positioned at one focus, and the shock wave generator at the other.

  
  

These examples highlight how the abstract concept of locus provides a powerful mathematical tool to understand, analyze, and innovate in the physical world around us. Mastering its principles offers a practical advantage far beyond exam halls.

Prerequisites for Locus and its Equation

Understanding the concept of Locus and deriving its equation is fundamental in Coordinate Geometry. Before delving into this topic, a strong foundation in several core mathematical concepts is essential. Mastering these prerequisites will not only make learning locus easier but also enhance your problem-solving speed and accuracy, crucial for both board exams and JEE.

Here are the key prerequisite concepts:

• Basic Algebra and Equation Solving:
  
  
  
  • Familiarity with algebraic manipulations, solving linear and quadratic equations, and systems of equations (two variables).
  
  
  • Understanding of inequalities and their graphical representation.
  
  
  • Relevance: Deriving locus equations often involves simplifying algebraic expressions and solving for relationships between 'x' and 'y'.
  
  
  
  


• Fundamental Concepts of Coordinate Geometry:
  
  
  
  • Cartesian Coordinate System: Understanding how points are represented in a 2D plane (x-axis, y-axis, origin).
  
  
  • Distance Formula: Calculating the distance between two points (x₁, y₁) and (x₂, y₂): $sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$.
  
  
  • Section Formula: Finding the coordinates of a point dividing a line segment in a given ratio (internal and external division), including the midpoint formula.
  
  
  • Area of a Triangle: Using coordinates to find the area of a triangle.
  
  
  • Relevance: Most locus problems define conditions based on distances, ratios, or positions of points, directly applying these formulas.
  
  
  
  


• Straight Lines:
  
  
  
  • Slope of a Line: Understanding how to calculate slope (m = (y₂-y₁)/(x₂-x₁)) and its geometric interpretation.
  
  
  • Different Forms of Line Equations:
    
    
    
    • Slope-intercept form: y = mx + c
    
    
    • Point-slope form: y - y₁ = m(x - x₁)
    
    
    • Two-point form: (y - y₁) = ((y₂-y₁)/(x₂-x₁))(x - x₁)
    
    
    • Intercept form: x/a + y/b = 1
    
    
    • General form: Ax + By + C = 0
    
    
    
    
  
  
  • Conditions for Parallel and Perpendicular Lines: $m_1 = m_2$ for parallel, $m_1 m_2 = -1$ for perpendicular.
  
  
  • Distance of a Point from a Line: Formula for distance from (x₁, y₁) to Ax + By + C = 0 is $|Ax_1 + By_1 + C| / sqrt{A^2 + B^2}$.
  
  
  • Relevance: Many locus conditions involve points equidistant from lines, or points lying on lines with specific slopes, making a thorough understanding of straight lines critical. For JEE Main, quick recall and application of these formulas are paramount.
  
  
  
  


• Geometric Properties and Definitions:
  
  
  
  • Basic knowledge of geometric shapes (triangle, square, circle, etc.) and their defining properties. For example, understanding what a perpendicular bisector is, or the definition of a circle.
  
  
  • Relevance: Locus is essentially the path of a point satisfying a given geometric condition. If you don't understand the condition itself (e.g., "equidistant from two points"), you can't translate it into an algebraic equation.
  
  
  
  

By ensuring proficiency in these areas, you will be well-equipped to tackle the challenges of Locus and its Equation, setting a strong base for subsequent topics like Conic Sections. Practice these fundamental concepts regularly to build confidence and speed.

Understanding Locus and its equation is a foundational skill in Coordinate Geometry. Many other essential topics are either prerequisites for or direct applications of locus concepts. Mastering these related areas will significantly strengthen your grasp of the subject.

Prerequisite Concepts

To effectively understand and solve problems related to locus, a solid understanding of the following basic coordinate geometry concepts is crucial:

• 
  Cartesian Coordinate System:
  
  
  
  • Understanding how points are represented by ordered pairs (x, y) in a 2D plane is the very first step.
  
  
  • Familiarity with quadrants and plotting points is fundamental to visualising any locus.
  
  
  
  


• 
  Distance Formula:
  
  
  
  • Many locus definitions involve conditions related to distances between points. The distance formula, $sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$, is extensively used to translate geometric conditions into algebraic equations.
  
  
  • JEE Relevance: It's a cornerstone for deriving equations of circles, parabolas, and other loci.
  
  
  
  


• 
  Section Formula:
  
  
  
  • This formula helps find the coordinates of a point that divides a line segment in a given ratio, internally or externally.
  
  
  • While less direct than the distance formula, it can be useful in specific locus problems involving midpoints or points dividing segments.
  
  
  
  


• 
  Basic Algebraic Equations:
  
  
  
  • Solving linear, quadratic, and sometimes higher-degree equations, along with manipulating algebraic expressions, is essential for simplifying and interpreting the locus equation once it's set up.
  
  
  • A strong command over algebraic manipulation ensures you can convert the geometric condition into its simplest algebraic form.
  
  
  
  

Direct Applications & Extensions

The concept of locus is not just a standalone topic; it is the fundamental principle behind the derivation and definition of many important geometric shapes.

• 
  Equation of a Straight Line:
  
  
  
  • A straight line can be defined as the locus of a point that is equidistant from two fixed points (perpendicular bisector) or a point that moves such that its slope with respect to a fixed point remains constant.
  
  
  • Understanding locus helps in deriving various forms of the line equation.
  
  
  
  


• 
  Equation of a Circle:
  
  
  
  • The definition of a circle is a classic example of locus: it's the locus of all points that are at a constant distance (radius) from a fixed point (centre).
  
  
  • Deriving the standard equation of a circle, $(x-h)^2 + (y-k)^2 = r^2$, directly uses this locus definition and the distance formula.
  
  
  
  


• 
  Conic Sections (Parabola, Ellipse, Hyperbola):
  
  
  
  • JEE/CBSE Relevance: This is where locus shines brightest in coordinate geometry. All three conic sections are *defined* as specific loci:
    
    
    
    • Parabola: Locus of points equidistant from a fixed point (focus) and a fixed line (directrix).
    
    
    • Ellipse: Locus of points such that the sum of their distances from two fixed points (foci) is constant.
    
    
    • Hyperbola: Locus of points such that the absolute difference of their distances from two fixed points (foci) is constant.
    
    
    
    
  
  
  • A deep understanding of locus is critical for deriving and understanding the properties of these curves.
  
  
  
  


• 
  Family of Curves:
  
  
  
  • While not a locus in itself, the concept of a "family" (e.g., family of lines passing through an intersection of two lines, or family of circles) often involves understanding how a curve's equation changes as a parameter varies, which is akin to tracing paths or conditions.
  
  
  
  

By clearly seeing how Locus connects to these other topics, you'll gain a more integrated and robust understanding of Coordinate Geometry, which is extremely beneficial for both board exams and competitive exams like JEE Main.

When dealing with Locus and its equation, students often fall into specific traps that can lead to incorrect answers. Being aware of these common pitfalls can significantly improve accuracy in both board exams and competitive exams like JEE Main.

Here are the common exam traps related to Locus and its equation:

• 
  Trap 1: Misinterpreting the Geometric Condition
  
  
  
  • Description: The most frequent error is incorrectly translating the verbal description of the locus into an algebraic condition. Students might confuse "distance from a point" with "distance from a line," or misunderstand terms like "equidistant," "sum of distances," or "ratio of distances."
  
  
  • JEE Relevance: JEE problems often use more complex or subtly worded conditions, demanding precise interpretation. For instance, "distance from the origin" is $sqrt{x^2+y^2}$, not just $x+y$.
  
  
  • Solution: Read the problem statement very carefully. Visualise the condition if possible. Break down complex conditions into simpler parts.
  
  
  
  


• 
  Trap 2: Ignoring Implicit Constraints or Domain Restrictions
  
  
  
  • Description: Locus problems can carry hidden conditions that restrict the valid range of $x$ or $y$. For example, if the locus involves square roots, the expression inside the root must be non-negative. If it involves the slope of a line segment, vertical lines (where slope is undefined) might need separate consideration or be excluded. Points where denominators become zero must also be excluded.
  
  
  • JEE Relevance: This is a major trap in JEE. For example, if the locus is defined by the midpoint of a chord of a circle, the midpoint must lie within the circle. If the condition states "distinct points," then certain equalities might need to be excluded.
  
  
  • Solution: Always consider what values of $x$ and $y$ are permissible based on the operations involved (square roots, division, logarithms) and the geometric context (e.g., inside a region, on a segment).
  
  
  
  


• 
  Trap 3: Algebraic Errors During Simplification
  
  
  
  • Description: Even with a correct setup, common algebraic mistakes during simplification can lead to an incorrect final equation. These include:
    
    
    
    • Errors in squaring expressions like $(a+b)^2$.
    
    
    • Incorrectly squaring both sides of an equation without considering extraneous solutions (especially when square roots are involved, ensure to check if original equation is satisfied).
    
    
    • Sign errors when rearranging terms.
    
    
    • Incorrectly multiplying or dividing terms.
    
    
    
    
  
  
  • CBSE Relevance: While CBSE focuses on the process, algebraic errors can still cost marks.
  
  
  • Solution: Work systematically, step-by-step. Double-check each algebraic manipulation. If squaring both sides, make sure to consider if the final equation introduces solutions that weren't part of the original equation's domain.
  
  
  
  


• 
  Trap 4: Not Recognizing the Standard Form of the Locus
  
  
  
  • Description: After deriving the equation, students sometimes fail to recognize it as a standard geometric shape (e.g., circle, parabola, ellipse, hyperbola, straight line). This can lead to missing parts of the question that ask for specific properties of the locus.
  
  
  • JEE Relevance: JEE questions frequently require identifying the type of conic section and its properties (eccentricity, foci, directrix, centre, radius).
  
  
  • Solution: Be very familiar with the general equations of all standard conic sections and how to transform a general second-degree equation ($Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$) into a standard form to identify its nature.
  
  
  
  


• 
  Trap 5: Assuming Specific Cases Instead of Generalizing
  
  
  
  • Description: Students might try to solve a locus problem by picking specific, convenient coordinates for the fixed points or lines (e.g., putting a point on an axis or at the origin). While this *can* sometimes give a hint, it's crucial to use general coordinates for the moving point $(x,y)$ and for any fixed points or lines unless their position is explicitly given.
  
  
  • JEE Relevance: Using specific cases might lead to a correct equation for that specific case but will fail for the general conditions of the problem.
  
  
  • Solution: Always use $(x,y)$ for the moving point whose locus is being found. Use general coordinates $(x_1, y_1)$, $(x_2, y_2)$ etc., for other fixed points or parameters as given, and only substitute specific values if explicitly stated.
  
  
  
  

By diligently checking for these common traps, you can significantly improve your accuracy and confidence in solving locus problems in exams.

Problem Solving Approach for Locus and its Equation

Understanding how to systematically approach locus problems is crucial for success in both board exams and competitive exams like JEE Main. Locus problems test your ability to translate geometric conditions into algebraic equations.

A locus is the path traced by a point that moves under one or more specified geometric conditions. Finding its equation involves a clear, step-by-step methodology.

General Steps to Determine the Locus

1. 
   Assume the Moving Point: Let the coordinates of the moving point, whose locus is to be found, be P(h, k). This is the starting point for almost all locus problems.
   


2. 
   Translate Geometric Conditions into Algebraic Relations: Express the given geometric conditions (e.g., distance, ratio of distances, angle, collinearity, perpendicularity, area) in terms of h, k, and the coordinates of any other fixed points or lines given in the problem.
   
   
   
   • For instance, if the distance from a point (x₁, y₁) is involved, use the distance formula: √[(h - x₁)² + (k - y₁)²].
   
   
   • If a point is equidistant from two lines, use the perpendicular distance formula.
   
   
   
   


3. 
   Formulate the Equation: Combine these algebraic relations to form an equation involving h and k. This is often the most critical step and might require careful algebraic manipulation.
   


4. 
   Eliminate Parameters (if any): Sometimes, the conditions might introduce auxiliary variables or parameters (e.g., slope 'm', an angle 'θ', a parameter 't'). If such parameters are introduced, they must be eliminated to obtain an equation solely in terms of h and k. This often involves using identities (like sin²θ + cos²θ = 1) or solving simultaneous equations.
   


5. 
   Replace (h, k) with (x, y): Once you have an equation involving only h and k, replace 'h' with 'x' and 'k' with 'y'. This gives the final equation of the locus.
   


6. 
   Simplify and Identify the Curve: Simplify the resulting equation to its standard form (e.g., Ax + By + C = 0 for a straight line, x² + y² + 2gx + 2fy + c = 0 for a circle, or the standard forms of parabola, ellipse, hyperbola). This helps in verifying the nature of the locus.
   

JEE vs. CBSE Approach:

While the fundamental steps remain the same, JEE problems often involve more intricate conditions, require more complex algebraic manipulation, and might lead to conic sections. CBSE problems are generally simpler, often resulting in straight lines or circles, and are more direct in their conditions.

Key Tips for JEE Main:

• 
  Geometric Interpretation: Always try to visualize the conditions. A quick sketch can often reveal the nature of the locus, helping you anticipate the form of the final equation.
  


• 
  Squaring Distances: When dealing with distances, it's often easier to work with the square of the distance to avoid square roots, provided all distances are positive.
  


• 
  Standard Equations: Be thorough with the standard equations of straight lines and all conic sections. This helps in quickly identifying the locus.
  


• 
  Exclusions: Sometimes, certain points might be excluded from the locus due to the conditions (e.g., a denominator becoming zero). Always check for such special cases.
  

Example:

Find the locus of a point P such that its distance from the point A(0,0) is twice its distance from the point B(3,0).

1. Let P be (h, k).


2. Given condition: PA = 2PB.
   Using the distance formula:
   PA = √[(h-0)² + (k-0)²] = √(h² + k²)
   PB = √[(h-3)² + (k-0)²] = √[(h-3)² + k²]
   


3. Substitute these into the condition:
   √(h² + k²) = 2 * √[(h-3)² + k²]
   Square both sides to eliminate square roots:
   h² + k² = 4 * [(h-3)² + k²]
   h² + k² = 4 * (h² - 6h + 9 + k²)
   h² + k² = 4h² - 24h + 36 + 4k²
   


4. Rearrange the terms to get an equation in h and k:
   3h² + 3k² - 24h + 36 = 0
   Divide by 3:
   h² + k² - 8h + 12 = 0
   


5. Replace (h, k) with (x, y):
   The equation of the locus is x² + y² - 8x + 12 = 0.
   


6. This is the equation of a circle. Completing the square:
   (x² - 8x + 16) + y² + 12 - 16 = 0
   (x - 4)² + y² = 4.
   This represents a circle with center (4,0) and radius 2.
   

Mastering this systematic approach will allow you to confidently tackle any locus problem in your exams!

CBSE Focus Areas: Locus and its Equation

In the CBSE curriculum, while 'Locus and its Equation' might not appear as an explicit, standalone chapter with complex problem types like in JEE, the underlying concept of locus is absolutely fundamental. It forms the basis for defining and deriving the equations of various standard geometric figures that are extensively covered in Class 11 and 12, such as lines, circles, parabolas, ellipses, and hyperbolas.

For CBSE board exams, the focus is primarily on understanding:

• The Definition of Locus: What exactly does it mean for a point to move under certain conditions?


• Basic Steps to Find a Locus Equation: How to translate geometric conditions into algebraic equations.


• Derivation of Standard Equations: Using locus definitions to derive the standard forms of equations for conic sections. This is a crucial aspect for board exams.

Key Concepts and Applications for CBSE:

You should be comfortable with applying the concept of locus in the following contexts:

1. 
   Perpendicular Bisector of a Line Segment:
   
   
   
   • The locus of a point equidistant from two fixed points (say A and B) is the perpendicular bisector of the line segment joining A and B.
   
   
   • You should be able to derive its equation.
   
   
   
   


2. 
   Circle:
   
   
   
   • The locus of a point that moves such that its distance from a fixed point (center) is always constant (radius).
   
   
   • Deriving the standard equation (x-h)² + (y-k)² = r² is a common and important application.
   
   
   
   


3. 
   Parabola:
   
   
   
   • The locus of a point that moves such that its distance from a fixed point (focus) is equal to its distance from a fixed line (directrix).
   
   
   • Derivation of standard forms like y² = 4ax or x² = 4ay is important.
   
   
   
   


4. 
   Ellipse:
   
   
   
   • The locus of a point such that the sum of its distances from two fixed points (foci) is a constant.
   
   
   • Understanding this definition helps in grasping the properties of the ellipse.
   
   
   
   


5. 
   Hyperbola:
   
   
   
   • The locus of a point such that the absolute difference of its distances from two fixed points (foci) is a constant.
   
   
   • Again, the definition itself is key for conceptual understanding.
   
   
   
   

CBSE vs. JEE Perspective:

Aspect	CBSE Board Exams	JEE Main / Advanced
Emphasis	Understanding the definition of locus and its direct application in deriving standard equations of lines and conic sections.	Solving complex problems where the locus itself is the primary unknown, often involving multiple conditions, transformations, or advanced geometric reasoning.
Problem Type	"Derive the equation of a circle...", "Find the equation of the locus of points equidistant from A and B."	"A point P moves such that...", often requiring algebraic manipulation beyond basic derivation, and sometimes involving calculus or other advanced concepts.
Complexity	Relatively straightforward, focusing on direct application of definitions.	High, requiring strong analytical skills and problem-solving strategies.

Tip for CBSE: Focus on the derivations of the standard equations of conic sections (circle, parabola, ellipse, hyperbola) from their definitions involving fixed points and lines. These derivations are often asked in board exams and directly test your understanding of locus.

Understanding the concept of Locus and its equation is fundamental in Coordinate Geometry. For JEE Main, this topic serves as a building block, often combined with concepts from straight lines, circles, and conic sections. Mastery here ensures a strong foundation for more complex problems.

JEE Focus Areas: Locus and its Equation

The core idea of finding the locus of a point is to define a path traced by a moving point (P) that satisfies certain geometric conditions. The equation of this path is the locus.

Key Principles & Strategies for JEE:

• Assume a Moving Point: Always start by assuming the coordinates of the moving point as P(h, k).


• Formulate the Condition Algebraically: Translate the given geometric condition into an algebraic equation involving 'h' and 'k'. This often involves using the distance formula, section formula, slope formula, or properties of geometric figures.


• Eliminate Parameters: If the moving point P(h, k) depends on some variable parameter (e.g., 't', 'm', 'θ'), eliminate this parameter from the equations to obtain a relation solely between 'h' and 'k'.


• Replace (h, k) with (x, y): Finally, replace 'h' with 'x' and 'k' with 'y' to get the equation of the locus in standard coordinate form.


• Identify the Curve: Be proficient in recognizing the type of curve (straight line, circle, parabola, ellipse, hyperbola) from its general equation.

Common JEE Problem Types:

1. 
   Distance-Based Loci:
   
   
   
   • Points equidistant from two fixed points (perpendicular bisector).
   
   
   • Points whose distance from a fixed point is constant (circle).
   
   
   • Points whose distances from two fixed points are in a constant ratio (Apollonius Circle).
   
   
   • Points such that the sum/difference of squares of distances from fixed points is constant.
   
   
   
   


2. 
   Geometric Property Loci:
   
   
   
   • Locus of the centroid, orthocentre, incentre, or circumcentre of a triangle whose vertices are variable but satisfy certain conditions (e.g., two vertices fixed, third vertex moves on a line/curve).
   
   
   • Locus of the foot of the perpendicular from a fixed point to a variable line.
   
   
   • Locus of the mid-point of a chord of a curve passing through a fixed point or parallel to a fixed direction.
   
   
   
   


3. 
   Parameter Elimination Loci:
   
   
   
   • When the coordinates (h, k) are expressed in terms of a single parameter (e.g., h = 2t+1, k = t-3). Eliminate 't' to find the locus.
   
   
   • Involving trigonometric identities to eliminate angles (e.g., h = a cos θ, k = b sin θ implies h^2/a^2 + k^2/b^2 = 1).
   
   
   
   

CBSE vs. JEE Main Perspective:

Aspect	CBSE Board Exam	JEE Main
Complexity	Generally straightforward, involving basic distance/ratio conditions.	More complex conditions, often requiring parameter elimination, advanced geometric properties, or combining concepts with conics.
Focus	Understanding the definition and basic equation derivation.	Problem-solving efficiency, identifying the resultant curve, and handling multiple variable conditions.
Skills	Algebraic manipulation, understanding definitions.	Strong analytical skills, geometric intuition, proficiency in parameter elimination, recognizing standard curve equations.

Common Pitfalls:

• Forgetting to eliminate the parameter.


• Making algebraic errors in distance formula or squaring expressions.


• Not considering restrictions on 'h' and 'k' that might arise from the initial conditions, leading to an incorrect domain for the locus.


• Incorrectly identifying the standard form of the derived equation.

Example: Find the locus of a point P such that the sum of the squares of its distances from two fixed points A(a, 0) and B(-a, 0) is equal to 2c².

Solution:

1. Let the moving point be P(h, k).


2. Given condition: PA² + PB² = 2c²


3. Using the distance formula:
   
   (h - a)² + (k - 0)² + (h - (-a))² + (k - 0)² = 2c²
   
   (h - a)² + k² + (h + a)² + k² = 2c²
   


4. Expand and simplify:
   
   h² - 2ah + a² + k² + h² + 2ah + a² + k² = 2c²
   
   2h² + 2k² + 2a² = 2c²
   


5. Divide by 2:
   
   h² + k² + a² = c²
   
   h² + k² = c² - a²
   


6. Replace (h, k) with (x, y):
   
   x² + y² = c² - a²
   

This is the equation of a circle with center (0, 0) and radius √(c² - a²), provided c² > a². If c² = a², the locus is a point (0,0). If c² < a², there is no real locus.

Mastering locus problems requires a blend of geometric intuition and strong algebraic manipulation skills. Practice diverse problems to build confidence!