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

Consider the hyperbola H: (x^2/(9)-(y^2/(16)=1.

One or more options correct

Consider the hyperbola

H:\ \dfrac{x^{2}}{9}-\dfrac{y^{2}}{16}=1

. Which of the following is (are) correct?

A. The eccentricity of $H$ is $\dfrac54$ .
B. For every point $P$ on $H$ , the product of the perpendicular distances of $P$ from the two asymptotes is $\dfrac{144}{25}$ .
C. The tangent to $H$ at the point $\left(5,\dfrac{16}{3}\right)$ is $5x-3y=9$ .
D. The line $y=2x+4$ is a tangent to $H$ .

Show answer and solution

Answer: B, C

Here

a=3

b=4

, so

c=5

. (A)

e=c/a=5/3

, not

5/4

: false. (B) The asymptotes are

4x\mp3y=0

. For

P(x_0,y_0)

the distances are

\frac{|4x_0-3y_0|}{5}

and

\frac{|4x_0+3y_0|}{5}

, with product

\frac{|16x_0^{2}-9y_0^{2}|}{25}=\frac{144\left|\frac{x_0^{2}}{9}-\frac{y_0^{2}}{16}\right|}{25}=\frac{144}{25}

: true. (C) The point lies on

H

since

\frac{25}{9}-\frac{256/9}{16}=1

. The tangent is

\frac{5x}{9}-\frac{(16/3)y}{16}=1

, i.e.

5x-3y=9

: true. (D) A line

y=mx+c_0

touches

H

iff

c_0^{2}=a^{2}m^{2}-b^{2}=36-16=20

; for

c_0=4

c_0^{2}=16\ne20

: false (the tangent would be

y=2x\pm2\sqrt5

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