JEE Advanced 2019 Paper 2 · Mathematics · Conic Sections · hard

(x^2/(25)+(y^2/(9)=1 with foci F1(-4,0) and F2(4,0).

One or more options correct

Consider the ellipse

E:\ \dfrac{x^{2}}{25}+\dfrac{y^{2}}{9}=1

with foci

F_1(-4,0)

and

F_2(4,0)

. Which of the following statements is (are) correct?

A. The eccentricity of $E$ is $\dfrac35$ .
B. The line $y=x+\sqrt{34}$ is a tangent to $E$ .
C. For every tangent to $E$ , the product of the perpendicular distances of $F_1$ and $F_2$ from the tangent equals $9$ .
D. The two tangents to $E$ drawn from the point $(5,3)$ are perpendicular to each other.

Show answer and solution

Answer: B, C, D

Here

a=5

b=3

c=4

. (A)

e=c/a=4/5

, not

3/5

: false. (B) A line

y=mx+c_0

touches the ellipse iff

c_0^{2}=a^{2}m^{2}+b^{2}

; with

m=1

c_0^{2}=25+9=34

: true. (C) For the tangent

y=mx+c_0

with

c_0^{2}=25m^{2}+9

, the distances of

(\pm4,0)

are

\frac{|\pm4m+c_0|}{\sqrt{1+m^{2}}}

; their product is

\frac{c_0^{2}-16m^{2}}{1+m^{2}}=\frac{9m^{2}+9}{1+m^{2}}=9=b^{2}

for every tangent: true. (D) The director circle is

x^{2}+y^{2}=a^{2}+b^{2}=34

, and

(5,3)

satisfies

25+9=34

, so the tangents from it are perpendicular: true.

Original question written in the JEE Advanced 2019 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