JEE Advanced 2022 Paper 2 · Mathematics · Conic Sections · medium

Let E be the ellipse (x^2/(16)+(y^2/(7)=1 with foci F1,F2.

One or more options correct

Let

E

be the ellipse

\dfrac{x^{2}}{16}+\dfrac{y^{2}}{7}=1

with foci

F_1,F_2

. Which of the following statements is/are correct?

A. The eccentricity of $E$ is $\dfrac34$
B. The minimum value of $PF_1\cdot PF_2$ over points $P$ on $E$ is $7$
C. The foot of the perpendicular from $F_1$ to any tangent of $E$ lies on the circle $x^{2}+y^{2}=7$
D. The length of the latus rectum of $E$ is $\dfrac74$

Show answer and solution

Answer: A, B

Here

a=4

b^2=7

, so

c^2=16-7=9

c=3

and

e=\frac34

; A is true. For

P=(x,y)

E

, the focal distances are

a\pm ex=4\pm\frac34x

, so

PF_1\cdot PF_2=16-\frac{9x^2}{16}

with

|x|\le4

; its minimum is at

x=\pm4

and equals

16-9=7

, so B is true. The foot of the perpendicular from a focus to a tangent lies on the auxiliary circle

x^2+y^2=a^2=16

, not

7

; C is false. The latus rectum has length

\frac{2b^2}{a}=\frac{14}4=\frac72

, not

\frac74

; D is false.

Original question written in the JEE Advanced 2022 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