Question

Tangent is drawn to ellipse $\frac{x^2}{27}+y^2=1$ at $(3\sqrt{3}\cos \theta ,\sin \theta )$ where $\theta \in (0,\frac{\pi }{2})$. Then the value of $\theta$ such that sum of intercepts on axes made by this tangent is minimum, is:

• A. $\frac{\pi }{3}$
• B. $\frac{\pi }{6}$
• C. $\frac{\pi }{8}$
• D. $\frac{\pi }{4}$

Answer 1

Answer: (B)

$\frac{\pi }{6}$

Given point is $\left(3\sqrt{3}\cos \theta ,\sin \theta \right)$

And equation of a tangent on a point (A,B) is :

$\frac{xA}{a^2}+\frac{yB}{b^2}=1$

Therefore we get, $\frac{x\cos \theta }{3\sqrt{3}}+\frac{y\sin \theta }{1}=1$

So, X-Intercept $X=\frac{3\sqrt{3}}{\cos \theta }$

and, Y-Intrcept $Y=\frac{1}{\sin \theta }$

Since sum of intercepts is minimum

therefore we have to minimize $f=\frac{3\sqrt{3}}{\cos \theta }+\frac{1}{\sin \theta }$

For minimum value $\frac{df}{d\theta }=0$

or, $\frac{3\sqrt{3}\sin \theta }{{\left(\cos \theta \right)}^2}-\frac{\cos \theta }{{\left(\sin \theta \right)}^2}=0$

On solving we get,

${\left(\tan \theta \right)}^3=3\sqrt{3}$

$or,\tan \theta =\sqrt{3}$

$\therefore \theta =\frac{\pi }{6}$