JEE Main 2023 — Ellipse Question with Solution

Given curve,

$$9x^2+16y^2=144$$

$$\Rightarrow \frac{x^2}{16}+\frac{y^2}{9}=1$$

$$\Rightarrow \frac{x^2}{4^2}+\frac{y^2}{3^2}=1$$

$$\therefore$$ a = 4 and b = 3

So, general point on the ellipse is $$=(4\cos \theta ,3\sin \theta )$$

We know,

Equation of tangent to a given ellipse at its point $$(a\cos \theta ,b\sin \theta )$$ is

$$\frac{x\cos \theta }{a}+\frac{y\sin \theta }{b}=1$$

$$\therefore$$ Here equation of tangent at point $$(4\cos \theta ,3\sin \theta )$$ is

$$\frac{x\cos \theta }{4}+\frac{y\sin \theta }{3}=1$$

When this tangent cut's x axis then y = 0.

$$\therefore$$ $$\frac{x\cos \theta }{4}+0=1$$

$$\Rightarrow x=4\sec \theta$$

$$\therefore$$ Point of intersection at x axis is $$A(4\sec \theta ,0)$$.

When this tangent cut's y axis then x = 0.

$$\therefore$$ $$0+\frac{y\sin \theta }{3}=1$$

$$\Rightarrow y=3\cos ec\theta$$

$$\therefore$$ Point of intersection at y axis is $$B(0,3\cos ec\theta )$$

$$\therefore$$ Length of AB

$$=\sqrt{{(4\sec \theta -0)}^2+{(0-3\cos ec\theta )}^2}$$

$$=\sqrt{16{\sec }^2\theta +9\cos e{c}^2\theta }$$

$$=\sqrt{16(1+{\tan }^2\theta )+9(1+{\cot }^2\theta )}$$

$$=\sqrt{25+16{\tan }^2\theta +9{\cot }^2\theta }$$

We know, $$AM\ge GM$$

$$\therefore$$ $$\frac{16{\tan }^2\theta +9{\cot }^2\theta }{2}\ge \sqrt{(16{\tan }^2\theta )(9{\cot }^2\theta )}$$

$$\Rightarrow 16{\tan }^2\theta +9{\cot }^2\theta \ge 2(4\tan \theta )(3\cot \theta )$$

$$\Rightarrow 16{\tan }^2\theta +9{\cot }^2\theta \ge 2\times 4\times 3$$

$$\Rightarrow 16{\tan }^2\theta +9{\cot }^2\theta \ge 24$$

$$\therefore$$ $$AB=\sqrt{25+16{\tan }^2\theta +9{\cot }^2\theta }$$

$$\ge \sqrt{25+24}$$

$$\ge \sqrt{49}$$

$$\ge 7$$

$$\therefore$$ Minimum length of $$AB=7$$.