(x^2/(16)+(y^2/(9)=1 with foci F1,F2.

One or more options correct

Consider the ellipse

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

with foci

F_1,F_2

. Which of the following is/are correct?

A. The minimum length of the portion of a tangent to $E$ intercepted between the coordinate axes is $7$
B. For any tangent to $E$ , the product of the perpendicular distances of $F_1$ and $F_2$ from it is $9$
C. The minimum area of the triangle formed by a tangent to $E$ and the coordinate axes is $24$
D. The tangent at the end $\left(\sqrt7,\tfrac94\right)$ of a latus rectum passes through $\left(\tfrac{16}{\sqrt7},0\right)$

Show answer and solution

Answer: A, B, D

For the point

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

the tangent meets the axes at

(4\sec\theta,0)

and

(0,3\csc\theta)

. Its intercepted length squared is

16\sec^2\theta+9\csc^2\theta\ge(4+3)^2=49

by Cauchy-Schwarz, so the minimum is

7

and (A) is true. The product of focal perpendiculars to a tangent equals

b^2=9

, which numerical checks at several points confirm, so (B) is true. The triangle area is

\tfrac12\cdot\dfrac{4}{\cos\theta}\cdot\dfrac{3}{\sin\theta}=\dfrac{6}{\sin\theta\cos\theta}=\dfrac{12}{\sin2\theta}\ge12

, so the minimum area is

12

, not

24

, and (C) is false. The tangent at

(\sqrt7,9/4)

\dfrac{\sqrt7x}{16}+\dfrac{y}{4}=1

, which meets the

x

-axis at

16/\sqrt7

, the foot of the directrix, so (D) is true.

Original question written in the JEE Advanced 2016 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