JEE Advanced 2018 Paper 1 · Mathematics · Conic Sections · medium

(x^2/(16)+(y^2/(9)=1 with foci S and S'.

One or more options correct

Consider the ellipse

E:\ \dfrac{x^2}{16}+\dfrac{y^2}{9}=1

with foci

S

and

S'

. Which of the following is (are) correct?

A. The minimum area of the triangle enclosed by a tangent to $E$ and the two coordinate axes is $12$
B. The product of the perpendicular distances from $S$ and $S'$ to any tangent of $E$ is $16$
C. The locus of the point from which two perpendicular tangents can be drawn to $E$ is a circle of radius $4$
D. The length of the latus rectum of $E$ is $9$

Show answer and solution

Answer: A

A tangent at

(4\cos\theta,3\sin\theta)

\frac{x\cos\theta}4+\frac{y\sin\theta}3=1

, with intercepts

\frac4{\cos\theta}

and

\frac3{\sin\theta}

. The triangle with the axes has area

\frac12\cdot\frac{12}{\sin\theta\cos\theta}=\frac{12}{\sin2\theta}\ge12

, with equality at

\theta=\frac\pi4

, so (A) is true. The product of the perpendicular distances from the two foci to any tangent equals

b^2=9

, not

16

, so (B) is false. The director circle is

x^2+y^2=a^2+b^2=25

, of radius

5

, not

4

, so (C) is false. The latus rectum has length

\frac{2b^2}{a}=\frac92

, not

9

, so (D) is false.

Original question written in the JEE Advanced 2018 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