Question

What is the minimum intercept made by the axes on the tangent to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ ?

• A. $a+b$
• B. $2(a+b)$
• C. $\sqrt{ab}$
• D. none of these

Answer 1

Answer: (A)

$a+b$

Equation of ellipse is$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
Let $P(a\cos \theta ,b\sin \theta )$ be a point on the ellipse.Equation of tangent to ellipse at P is $\frac{x}{a}\cos \theta +\frac{y}{b}\sin \theta =1$
So, the tangent intersects X-axis at $(a\sec \theta ,0)$ and Y-axis at $(0,bcosec\theta )$
Intercepted portion of tangent between axes is $S=\sqrt{a^2{\sec }^2\theta +b^{2}cose{c}^2\theta }$
For maximum or minimum , $\frac{dS}{d\theta }=0$$\Rightarrow 2a^2{\sec }^2\theta \tan \theta -2b^{2}cose{c}^2\theta \cot \theta =0$$\Rightarrow {\tan }^4\theta =\frac{b^2}{a^2}$$\Rightarrow {\tan }^2\theta =\frac{b}{a}$
Also, $\frac{d^{2}S}{d{\theta }^2}>0$ for ${\tan }^2\theta =\frac{b}{a}$
Hence, S attains minimum at ${\tan }^2\theta =\frac{b}{a}$
Minimum value of $S=\sqrt{a^2(1+{\tan }^2\theta )+b^2(1+{\cot }^2\theta )}$$\Rightarrow S_{min}=\sqrt{a(a+b)+b(a+b)}$
Minimum value of S $=a+b$