JEE Main 2020 · Mathematics · Conic Sections · hard

A tangent to the ellipse (x^2/(25)+(y^2/(9)=1 at a point in the first quadrant meets the coordinate axes at A and B. The minimum possible length of…

Single correct option

A tangent to the ellipse

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

at a point in the first quadrant meets the coordinate axes at

A

and

B

. The minimum possible length of

AB

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Answer: (A) $8$

The tangent at

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

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

, with intercepts

A\left(\dfrac5{\cos\theta},0\right)

and

B\left(0,\dfrac3{\sin\theta}\right)

. Then

AB^2=\frac{25}{\cos^2\theta}+\frac{9}{\sin^2\theta}.

By Cauchy-Schwarz,

\left(\dfrac{25}{c^2}+\dfrac9{s^2}\right)(c^2+s^2)\ge(5+3)^2

, and

c^2+s^2=1

, so

AB^2\ge64

. Equality holds when

\tan^2\theta=\dfrac35

, which is attainable in the first quadrant. Hence the minimum length is

5+3=8

. So the tangent segment is shortest at that point, and its length equals the sum of the semi-axes.

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