JEE Main 2025 (January)

LEVELJEE Advanced

Animated Solution for Mathematics - Conic Sections: If A and B are the points of intersection of the circle $x^{2} + y^{2} - 8 x = 0$ and the hyperbola $\frac{x ^{2}}{9} - \frac{y ^{2}}{4} = 1$ and a point P moves on the line $2 x - 3 y + 4 = 0$ then the centroid of $△ P A B$ lies on the line :

Q: Why did we reject the negative root for $x$?

Great observation. When we solved the quadratic, we obtained $x = -\frac{6}{13}$. If you plug this into the circle equation $y^2 = 8x - x^2$, you get a negative value for $y^2$. Since $y$ must be a real number for the intersection point to exist on the Cartesian plane, we must discard this root. Always check the physical validity of your algebraic results!

Q: Do I really need to find the exact coordinates of $A$ and $B$?

In this specific problem, yes, because the centroid formula depends on the specific values of the vertices. However, notice how the $y$-coordinates of $A$ and $B$ were additive inverses ($2\sqrt{3}$ and $-2\sqrt{3}$)? They cancelled out perfectly in the centroid formula. Sometimes, you don't need the exact values, just their sum or relationship. Keep an eye out for that!

Q: How do I know when to replace $(h, k)$ with $(x, y)$?

The variables $(h, k)$ represent a specific point on the locus. To define the *entire* path (the locus) as a general equation, we replace the specific coordinates $(h, k)$ with the general variables $(x, y)$. It's like moving from a snapshot of a single point to the equation of the entire trail it leaves behind.

Select Answer:

Visualized Solution

System of Equations

Circle: $x^{2} + y^{2} - 8 x = 0$
Hyperbola: $\frac{x ^{2}}{9} - \frac{y ^{2}}{4} = 1$
Goal: Find intersection points $A$ and $B$ .

Substitution Method

From circle: $y^{2} = 8 x - x^{2}$
Substitute into hyperbola: $\frac{x ^{2}}{9} - \frac{8 x - x ^{2}}{4} = 1$

Algebraic Simplification

Multiply by $36$ (LCM of $9$ and $4$ ): $4 x^{2} - 9 (8 x - x^{2}) = 36$
Expand brackets: $4 x^{2} - 72 x + 9 x^{2} = 36$
Combine terms: $13 x^{2} - 72 x - 36 = 0$

Solving the Quadratic

Factorize: $13 x^{2} - 78 x + 6 x - 36 = 0$
Group terms: $13 x (x - 6) + 6 (x - 6) = 0$
Roots: $(13 x + 6) (x - 6) = 0$
Possible values: $x = 6$ or $x = - \frac{6}{13}$

Coordinates of A and B

For $x = - \frac{6}{13}$ , $y^{2} = 8 (- \frac{6}{13}) - (- \frac{6}{13})^{2} < 0$ (Rejected)
For $x = 6$ , $y^{2} = 8 (6) - 6^{2} = 48 - 36 = 12$
$y = \pm 12 = \pm 23$
Points: $A (6, 23)$ and $B (6, - 23)$

Locus Setup

Point $P (x_{1}, y_{1})$ moves on line: $2 x - 3 y + 4 = 0$
Centroid $G (h, k)$ of $△ P A B$
Centroid Formula: $h = \frac{x _{1} + x _{2} + x _{3}}{3}, k = \frac{y _{1} + y _{2} + y _{3}}{3}$

Applying Centroid Formula

$x$ -coordinate: $h = \frac{6 + 6 + x _{1}}{3} \Rightarrow 3 h = 12 + x_{1}$
$y$ -coordinate: $k = \frac{2 3 - 2 3 + y _{1}}{3} \Rightarrow 3 k = y_{1}$
Express $P$ : $x_{1} = 3 h - 12$ and $y_{1} = 3 k$

Locus Condition

Since $P (x_{1}, y_{1})$ lies on $2 x - 3 y + 4 = 0$
Substitute $x_{1}$ and $y_{1}$ :
$2 (3 h - 12) - 3 (3 k) + 4 = 0$

Final Equation

Expand: $6 h - 24 - 9 k + 4 = 0$
Simplify: $6 h - 9 k - 20 = 0 \Rightarrow 6 h - 9 k = 20$
Replace $(h, k)$ with $(x, y)$ :
Locus: $6 x - 9 y = 20$

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The Sigma Insight: Standard Equations of Parabola, Ellipse, and Hyperbola

Solution Diagram

Analyzing the Intersection

To begin, we identify the intersection points

A

and

B

of the circle

x^{2} + y^{2} - 8 x = 0

and the hyperbola

\frac{x ^{2}}{9} - \frac{y ^{2}}{4} = 1

. We isolate

y^{2}

from the circle equation:

y^{2} = 8 x - x^{2}

Substituting this into the hyperbola equation, we obtain:

\frac{x ^{2}}{9} - \frac{8 x - x ^{2}}{4} = 1

Multiplying the entire equation by

36

to clear the denominators, we get

4 x^{2} - 9 (8 x - x^{2}) = 36

. Expanding and rearranging terms leads to the quadratic equation:

13 x^{2} - 72 x - 36 = 0

Factoring this expression, we find

(13 x + 6) (x - 6) = 0

. This yields two potential values for

x

x = 6

and

x = - \frac{6}{13}

Validating the Intersection Points

We must verify these roots against the circle equation. For

x = - \frac{6}{13}

, the term

y^{2} = 8 (- \frac{6}{13}) - (- \frac{6}{13})^{2}

results in a negative value, which is impossible for real coordinates.

Thus, we accept only

x = 6

. Substituting

x = 6

back into the circle equation gives

y^{2} = 8 (6) - 36 = 12

, so

y = \pm 23

. The intersection points are

A (6, 23)

and

B (6, - 23)

The Geometry of the Centroid

Let

P (x_{1}, y_{1})

be a point on the line

2 x - 3 y + 4 = 0

. The centroid

G (h, k)

△ P A B

is calculated as the average of the vertices:

h = \frac{x _{1} + 6 + 6}{3}, k = \frac{y _{1} + 2 3 - 2 3}{3}

Simplifying these expressions, we find:

3 h = x_{1} + 12 ⟹ x_{1} = 3 h - 12

3 k = y_{1} ⟹ y_{1} = 3 k

Determining the Locus

Since

P (x_{1}, y_{1})

lies on the line

2 x - 3 y + 4 = 0

, we substitute our expressions for

x_{1}

and

y_{1}

into this constraint:

2 (3 h - 12) - 3 (3 k) + 4 = 0

Expanding the terms, we get

6 h - 24 - 9 k + 4 = 0

. This simplifies to

6 h - 9 k - 20 = 0

Replacing

(h, k)

with the general coordinates

(x, y)

, the locus of the centroid is:

6x - 9y = 20

Common Traps & FAQs

Why did we reject the negative root for

x

Great observation. When we solved the quadratic, we obtained

x = - \frac{6}{13}

. If you plug this into the circle equation

y^{2} = 8 x - x^{2}

, you get a negative value for

y^{2}

. Since

y

must be a real number for the intersection point to exist on the Cartesian plane, we must discard this root. Always check the physical validity of your algebraic results!

Do I really need to find the exact coordinates of

A

and

B

In this specific problem, yes, because the centroid formula depends on the specific values of the vertices. However, notice how the

y

-coordinates of

A

and

B

were additive inverses (

23

and

- 23

)? They cancelled out perfectly in the centroid formula. Sometimes, you don't need the exact values, just their sum or relationship. Keep an eye out for that!

How do I know when to replace

(h, k)

with

(x, y)

The variables

(h, k)

represent a specific point on the locus. To define the *entire* path (the locus) as a general equation, we replace the specific coordinates

(h, k)

with the general variables

(x, y)

. It's like moving from a snapshot of a single point to the equation of the entire trail it leaves behind.

Animated Solution for Mathematics - Conic Sections: If A and B are the points of intersection of the circle $x^{2} + y^{2} - 8 x = 0$ and the hyperbola $\frac{x ^{2}}{9} - \frac{y ^{2}}{4} = 1$ and a point P moves on the line $2 x - 3 y + 4 = 0$ then the centroid of $△ P A B$ lies on the line :

Select Answer:

Visualized Solution

System of Equations

Substitution Method

Algebraic Simplification

Solving the Quadratic

Coordinates of A and B

Locus Setup

Applying Centroid Formula

Locus Condition

Final Equation

The Sigma Insight: Standard Equations of Parabola, Ellipse, and Hyperbola

Analyzing the Intersection

Validating the Intersection Points

The Geometry of the Centroid

Determining the Locus

Similar Questions

The locus of the centroid of the triangle formed by any point $P$ on the hyperbola $16 x^{2} - 9 y^{2} + 32 x + 36 y - 164 = 0$ , and its foci is :

Let $A$ be the focus of the parabola $y^{2} = 8 x$ . Let the line $y = m x + c$ intersect the parabola at two distinct points $B$ and $C$ . If the centroid of the triangle $A B C$ is $(\frac{7}{3}, \frac{4}{3})$ , then $(B C)^{2}$ is equal to :

Let $y^{2} = 12 x$ be the parabola with its vertex at $O$ . Let $P$ be a point on the parabola and $A$ be a point on the $x$ -axis such that $∠ O P A = 9 0^{\circ}$ . Then the locus of the centroid of such triangles $O P A$ is :

Comprehension Passage

Equation of a common tangent with positive slope to the circle as well as to the hyperbola is

Equation of the circle with $A B$ as its diameter is

Tangents are drawn from any point on the hyperbola $\frac{x ^{2}}{9} - \frac{y ^{2}}{4} = 1$ to the circle $x^{2} + y^{2} = 9$ . Find the locus of mid-point of the chord of contact.

Let $S$ be the focus of the parabola $y^{2} = 8 x$ and let $P Q$ be the common chord of the circle $x^{2} + y^{2} - 2 x - 4 y = 0$ and the given parabola. The area of the triangle $P QS$ is

Let P be a point on the hyperbola $H : \frac{x ^{2}}{9} - \frac{y ^{2}}{4} = 1$ , in the first quadrant such that the area of triangle formed by P and the two foci of H is $213$ . Then, the square of the distance of P from the origin is

Let $P$ be the point $(1, 0)$ and $Q$ a point on the locus $y^{2} = 8 x$ . The locus of mid point of $P Q$ is

Comprehension Passage

The ratio of the areas of the triangles $P QS$ and $P QR$ is

The radius of the circumcircle of the triangle $P R S$ is

The radius of the incircle of the triangle $P QR$ is

Let $R$ be the focus of the parabola $y^{2} = 20 x$ and the line $y = m x + c$ intersect the parabola at two points $P$ and $Q$ . Let the points $G (10, 10)$ be the centroid of the triangle $P QR$ . If $c - m = 6$ , then $P Q^{2}$ is

Select Answer:

Visualized Solution

System of Equations

Substitution Method

Algebraic Simplification

Solving the Quadratic

Coordinates of A and B

Locus Setup

Applying Centroid Formula

Locus Condition

Final Equation

The Sigma Insight: Standard Equations of Parabola, Ellipse, and Hyperbola

Analyzing the Intersection

Validating the Intersection Points

The Geometry of the Centroid

Determining the Locus

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The locus of the centroid of the triangle formed by any point P on the hyperbola 16x2−9y2+32x+36y−164=0, and its foci is :

Comprehension Passage

Equation of a common tangent with positive slope to the circle as well as to the hyperbola is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equation of the circle with AB as its diameter is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Tangents are drawn from any point on the hyperbola 9x2​−4y2​=1 to the circle x2+y2=9. Find the locus of mid-point of the chord of contact.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let S be the focus of the parabola y2=8x and let PQ be the common chord of the circle x2+y2−2x−4y=0 and the given parabola. The area of the triangle PQS is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let P be the point (1,0) and Q a point on the locus y2=8x. The locus of mid point of PQ is

Comprehension Passage

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The ratio of the areas of the triangles PQS and PQR is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The radius of the circumcircle of the triangle PRS is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The radius of the incircle of the triangle PQR is

The locus of the centroid of the triangle formed by any point $P$ on the hyperbola $16 x^{2} - 9 y^{2} + 32 x + 36 y - 164 = 0$ , and its foci is :

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Tangents are drawn from any point on the hyperbola $\frac{x ^{2}}{9} - \frac{y ^{2}}{4} = 1$ to the circle $x^{2} + y^{2} = 9$ . Find the locus of mid-point of the chord of contact.

Let $S$ be the focus of the parabola $y^{2} = 8 x$ and let $P Q$ be the common chord of the circle $x^{2} + y^{2} - 2 x - 4 y = 0$ and the given parabola. The area of the triangle $P QS$ is

Let $P$ be the point $(1, 0)$ and $Q$ a point on the locus $y^{2} = 8 x$ . The locus of mid point of $P Q$ is

The ratio of the areas of the triangles $P QS$ and $P QR$ is

The radius of the circumcircle of the triangle $P R S$ is

The radius of the incircle of the triangle $P QR$ is