JEE Advanced 2025 Paper 2 · Mathematics · Conic Sections · hard

A tangent at a point P in the first quadrant of the ellipse (x^2/(25)+(y^2/(16)=1 meets the coordinate axes at A and B. When the length AB is…

Numerical answer

A tangent at a point

P

in the first quadrant of the ellipse

\dfrac{x^2}{25}+\dfrac{y^2}{16}=1

meets the coordinate axes at

A

and

B

. When the length

AB

is minimum, the

x

-coordinate of

P

is ______. Give the answer correct to two decimal places.

Show answer and solution

Answer: 3.73

Take

P=(5\cos\theta,4\sin\theta)

; the tangent is

\frac{x\cos\theta}{5}+\frac{y\sin\theta}{4}=1

, with intercepts

A=(5\sec\theta,0)

B=(0,4\csc\theta)

. Then

AB^2=25\sec^2\theta+16\csc^2\theta

. Setting the derivative to zero:

25\sec^2\theta\tan\theta=16\csc^2\theta\cot\theta

, so

25\sin^4\theta=16\cos^4\theta

, i.e.

\tan^2\theta=\frac45

. Then

\cos^2\theta=\frac{1}{1+4/5}=\frac59

\cos\theta=\frac{\sqrt5}{3}

. The minimum

AB

5+4=9

, and

x_P=5\cos\theta=\frac{5\sqrt5}{3}\approx3.73

(matches numerical minimisation).

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