JEE Advanced 2023 Paper 1 · Mathematics · Conic Sections · medium

A variable tangent to the ellipse (x^2/(25)+(y^2/(9)=1 at a point in the first quadrant meets the x-axis at A and the y-axis at B. The minimum…

Single correct option

A variable tangent to the ellipse

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

at a point in the first quadrant meets the

x

-axis at

A

and the

y

-axis at

B

. The minimum possible length of the segment

AB

Show answer and solution

Answer: (C) $8$

Take the point

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

with

0<\theta<\frac\pi2

. The tangent is

\frac{x\cos\theta}{5}+\frac{y\sin\theta}{3}=1

, which meets the axes at

A\left(\frac5{\cos\theta},0\right)

and

B\left(0,\frac3{\sin\theta}\right)

. Then

AB^2=25\sec^2\theta+9\csc^2\theta=25+25\tan^2\theta+9+9\cot^2\theta=34+25\tan^2\theta+9\cot^2\theta

. By AM-GM,

25\tan^2\theta+9\cot^2\theta\ge2\sqrt{225}=30

, with equality when

\tan^2\theta=\frac35

. So

AB^2_{\min}=64

and

AB_{\min}=8=a+b

. The value

\sqrt{34}

comes from dropping the term

25\tan^2\theta+9\cot^2\theta

altogether,

5

is only the semi-major axis, and

2\sqrt{34}

is an arbitrary overestimate; none of them is attained by a tangent.

Original question written in the JEE Advanced 2023 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