If A and B are the points of intersection of the circle x^2+y^2-8 x=0 and the hyperbola (x^2)/(9)-(y^2)/(4)=1 and a point P moves on the line 2 x-3 y+4=0 , then the centroid of PAB lies on the line :

Correct answer: C. 6 x-9 y=20 C: x^2+y^2-8 x=0 H: (x^2)/(9)-(y^2)/(4)=1 By solving (x^2)/(9)-((8 x-x^2)/(4))=1 aligned & 4 x^2-72 x+9 x^2=36 & 13 x^2-72 x-36=0 & 13 x^2-78 x+6 x-36=0 & 13 x(x-6)+6(x-6)=0 & x=6 or -(13)/(6) × neglected & y^2=8(6)-(6)^2 & y= ± 12 aligned So, points A and B are (6, 12 ),(6,- 12 ) P(h, (2 h+4)/(3)) Centroid of P A B is ((12+h)/(3), (2 h+4)/(9)) By options this centroid lies on the live 6 x-9 y=20

JEE Main 202528 JanuaryEvening Shift

Exam: JEE Main
Year: 2025
Date: 28 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Hyperbola

JEE Main 2025 · 28 January · Evening Shift · Mathematics · Hyperbola

If A and B are the points of intersection of the circle x^2+y^2-8...

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

\triangle \mathrm{PAB}

lies on the line :

Options

A $x+9 y=36$
B $4 x-9 y=12$
C $6 x-9 y=20$
D $9 x-9 y=32$

Correct answer

C. $6 x-9 y=20$

Step-by-step solution

x^2+y^2-8 x=0 \mathrm{H}: \frac{x^2}{9}-\frac{y^2}{4}=1

By solving

\frac{x^2}{9}-\left(\frac{8 x-x^2}{4}\right)=1

\begin{aligned} & 4 x^2-72 x+9 x^2=36 \\ & \Rightarrow 13 x^2-72 x-36=0 \\ & \Rightarrow 13 x^2-78 x+6 x-36=0 \\ & \Rightarrow 13 x(x-6)+6(x-6)=0 \\ & \Rightarrow x=6 \text{\text{ or } }-\frac{13}{6} \times \text{ neglected } \\ & \Rightarrow y^2=8(6)-(6)^2 \\ & \Rightarrow y= \pm \sqrt{12} \end{aligned}

So, points

A and B

are

(6, \sqrt{12}),(6,-\sqrt{12}) P\left(h, \frac{2 h+4}{3}\right)

Centroid of

\triangle P A B

is \left(\frac{12+h}{3}, \frac{2 h+4}{9}\right)

By options this centroid lies on the live $6 x-9 y=20

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