JEE Advanced 2020 Paper 1 · Mathematics · Conic Sections · medium

Consider the ellipse E:(x^2/(9)+(y^2/(4)=1. The tangent at the point P(3cos θ,2sin θ), 0<θ<π/2, meets the coordinate axes at A and B.

One or more options correct

Consider the ellipse

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

. The tangent at the point

P(3\cos\theta,2\sin\theta)

0<\theta<\pi/2

, meets the coordinate axes at

A

and

B

. Which of the following is/are correct?

A. For $\theta=\pi/3$ , the area of triangle $OAB$ ( $O$ the origin) is $4\sqrt3$
B. The minimum possible area of triangle $OAB$ over all such $\theta$ is $3$
C. The locus of points from which two perpendicular tangents can be drawn to $E$ is $x^2+y^2=5$
D. The foci of $E$ are $(\pm\sqrt{13},0)$

Show answer and solution

Answer: A

The tangent at

P

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

, with intercepts

A(3/\cos\theta,0)

and

B(0,2/\sin\theta)

. The area is

\tfrac12\cdot\tfrac{3}{\cos\theta}\cdot\tfrac{2}{\sin\theta}=\dfrac{6}{\sin2\theta}

. (A) At

\theta=\pi/3

\sin2\theta=\sqrt3/2

, so the area is

12/\sqrt3=4\sqrt3

; true. (B) Since

\sin2\theta\le1

, the area is at least

6

(attained at

\theta=\pi/4

), not

3

; false. (C) The director circle of

x^2/a^2+y^2/b^2=1

x^2+y^2=a^2+b^2=13

, not

5

; false. (D) With

c^2=a^2-b^2=9-4=5

, the foci are

(\pm\sqrt5,0)

;

\sqrt{13}

comes from wrongly adding the squares. Only (A) is correct.

Original question written in the JEE Advanced 2020 Paper 1 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake