Question

Tangents are drawn to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, $(a>b)$, and the circle $x^2+y^2=a^2$ at the points where a common ordinate cuts them (on the same side of the $x$-axis). Then the greatest acute angle between these tangents is given by

• A. ${\tan }^{-1}\left(\frac{a-b}{2\sqrt{ab}}\right)$
• B. ${\tan }^{-1}\left(\frac{a+b}{2\sqrt{ab}}\right)$
• C. ${\tan }^{-1}\left(\frac{2ab}{\sqrt{a-b}}\right)$
• D. ${\tan }^{-1}\left(\frac{2ab}{\sqrt{a+b}}\right)$

Answer 1

Answer: (A)

Tangent to the ellipse at $P(a\cos \alpha ,b\sin \alpha )$ is
$\frac{x}{a}\cos \alpha +\frac{y}{b}\sin \alpha =\mathrm{1...}$(i)
Tangent to the circle at $Q(a\cos \alpha ,a\sin \alpha )$ is
$\cos \alpha x+\sin \alpha y=a....$ (ii)
Now, the angle between the tangents is $\theta$. Then,
$\tan \theta =\left|\frac{-\frac{b}{a}\cot \alpha -(-\cot \alpha )}{1+\left(-\frac{b}{a}\cot \alpha \right)(-\cot \alpha )}\right|$
$=\left|\frac{\cot \alpha \left(1-\frac{b}{a}\right)}{|1+\frac{b}{a}{\cot }^2\alpha |}\right|=\left|\frac{a-b}{a\tan \alpha +b\cot \alpha }\right|$
$=\left|\frac{a-b}{(\sqrt{a\tan \alpha }-\sqrt{b\cot \alpha }{)}^2+2\sqrt{ab}}\right|$
Now, the greatest value of the above expression is $\left|\frac{a-b}{2\sqrt{ab}}\right|$ when $\sqrt{a\tan \alpha }=\sqrt{b\cot \alpha }.$ Therefore,
${\theta }_{\text{maximum }}={\tan }^{-1}\left(\frac{a-b}{2\sqrt{ab}}\right)$