Question

Let $E$ be an standard ellipse (center at the origin and major axis as $x-$axis) whose length of major axis is $2a$ and length of minor axis is $2b$. Let $C$ be a circle with centre at origin and cutting the ellipse at an angle $\alpha$, then the value of radius of $C$ for which the angle $\alpha$ is maximum is

• A. $\sqrt{\frac{a^2-b^2}{2}}$
• B. $\sqrt{\frac{a^2+b^2}{2}}$
• C. $\sqrt{a^2-b^2}$
• D. $\frac{1}{2}\sqrt{a^2-b^2}$

Answer 1

The correct option is B $\sqrt{\frac{a^2+b^2}{2}}$


Equation of the ellipse is given by,
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
Assuming a $p=(a\cos \theta ,b\sin \theta )$
Equation of the tangent at $p$ on the ellipse:
$\frac{x\times a\cos \theta }{a^2}+\frac{y\times b\sin \theta }{b^2}=1\Rightarrow y=\frac{b}{\sin \theta }(1-\frac{x\cos \theta }{a})$
Slope of the tangent:
$m_1=\tan \beta =-\frac{b\cos \theta }{a\sin \theta }$
Equation of tangent at $p$ on the circle:
$x\times a\cos \theta +y\times b\sin \theta =r^2$
$m_2=\tan \gamma =-\frac{a\cos \theta }{b\sin \theta }$
So the angle between the curve can be written as,
$\tan \alpha =|\frac{m_1-m_2}{1+m_1{m}_2}|\Rightarrow \tan \alpha =\left|\frac{-\frac{b\cos \theta }{a\sin \theta }+\frac{a\cos \theta }{b\sin \theta }}{1+\frac{ab{\cos }^2\theta }{ab{\sin }^2\theta }}\right|\Rightarrow \tan \alpha =\frac{\sin \theta \cos \theta (a^2-b^2)}{ab}\Rightarrow \tan \alpha =\frac{\sin 2\theta (a^2-b^2)}{2ab}$
So $\alpha$ will be maximum when $\sin 2\theta =1\Rightarrow \theta =\frac{\pi }{4}$
Now we can write radius of the circle as,
$r=\sqrt{a^2{\cos }^2\theta +b^2{\sin }^2\theta }=\sqrt{\frac{a^2+b^2}{2}}$