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

Answer 1

Answer: (A)

Tangent to the ellipse at $P ( a \cos \alpha, b \sin \alpha)$ is
$\frac{x}{a} \cos \alpha+\frac{y}{b} \sin \alpha=1 ...$(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)}{\left|1+\frac{b}{a} \cot ^{2} \alpha\right|}\right|=\left|\frac{a-b}{a \tan \alpha+b \cot \alpha}\right| $
$=\left|\frac{a-b}{(\sqrt{a \tan \alpha}-\sqrt{b \cot \alpha})^{2}+2 \sqrt{a b}}\right|$
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,
$\theta_{\text {maximum }}=\tan ^{-1}\left(\frac{a-b}{2 \sqrt{a b}}\right)$