JEE Advanced 2024 Paper 2 · Mathematics · Conic Sections · medium

The tangent at a point P in the first quadrant on the ellipse (x^2/(9)+(y^2/(4)=1 meets the coordinate axes at A and B. When the area of triangle OAB…

Single correct option

The tangent at a point

P

in the first quadrant on the ellipse

\dfrac{x^2}{9}+\dfrac{y^2}{4}=1

meets the coordinate axes at

A

and

B

. When the area of triangle

OAB

(

O

the origin) is as small as possible, the square of the radius of the circle passing through

O

A

and

B

equals

Show answer and solution

Answer: (D) $\dfrac{13}{2}$

Write

P=(3\cos\phi,2\sin\phi)

. The tangent is

\dfrac{x\cos\phi}{3}+\dfrac{y\sin\phi}{2}=1

, giving intercepts

OA=\dfrac{3}{\cos\phi}

and

OB=\dfrac{2}{\sin\phi}

. The area is

\dfrac12\cdot\dfrac{6}{\sin\phi\cos\phi}=\dfrac{6}{\sin2\phi}

, minimum

6

\phi=\pi/4

. Then

OA=3\sqrt2

OB=2\sqrt2

and

AB^2=18+8=26

. The angle at

O

is a right angle, so

AB

is a diameter of the circle through

O,A,B

and

R^2=AB^2/4=26/4=13/2

. The value

13/4

takes

AB/2

with a wrong

AB^2=13

;

13

AB^2/2

;

9/2

uses only

OA

Original question written in the JEE Advanced 2024 Paper 2 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake