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 least is

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

Answer 1

The correct option is B $\frac{\pi }{6}$

Let $x=3\sqrt{3}\cos \theta$
And
$y=\sin \theta$
Hence, $\frac{dy}{dx}$
$=-\frac{\cos \theta }{3\sqrt{3}\sin \theta }$ ...(i)
Hence equation of the tangent will be
$y-\sin \theta =-\frac{\cos \theta }{3\sqrt{3}\sin \theta }(x-3\sqrt{3}.\cos \theta )$
$3\sqrt{3}\sin \theta .y-3\sqrt{3}{\sin }^2\theta =-x\cos \theta +3\sqrt{3}{\cos }^2\theta$
$3\sqrt{3}\sin \theta .y+x\cos \theta =3\sqrt{3}$.
Hence the intercepts are
$cosec\theta$ and $3\sqrt{3}.\sec \theta$.
Hence, $k=cosec\theta +3\sqrt{3}.\sec \theta$
Now $\frac{dk}{d\theta }$ $=-cosec\theta .\cot \theta +3\sqrt{3}.\sec \theta .\tan \theta$ $=0$
$3\sqrt{3}.\sec \theta .\tan \theta =cosec\theta .\cot \theta$
$3\sqrt{3}.{\tan }^2\theta .\frac{1}{\cos \theta }=\frac{1}{\sin \theta }$
$3\sqrt{3}.{\tan }^2\theta =\frac{\cos \theta }{\sin \theta }$
$3\sqrt{3}.{\tan }^3\theta =1$
${\tan }^3\theta =\frac{1}{3\sqrt{3}}$
$\tan \theta =\frac{1}{\sqrt{3}}$
$\theta ={\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right)$
$\theta ={30}^0$ $=\frac{\pi }{6}$.