Question

A circle of radius $r$ is concentric with the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$. Find the angle made by the common tangent with the axis of the ellipse.

• A. $ta{n}^{-1}\sqrt{\frac{(r^2-a^2)}{(b^2-r^2)}}$
• B. $ta{n}^{-1}\sqrt{\frac{(r-b)}{(a-r)}}$
• C. $ta{n}^{-1}\sqrt{\frac{(b^2-r^2)}{(a^2-r^2)}}$
• D. $ta{n}^{-1}\sqrt{\frac{(r^2-b^2)}{(a^2-r^2)}}$

Answer 1

The correct option is D $ta{n}^{-1}\sqrt{\frac{(r^2-b^2)}{(a^2-r^2)}}$

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

$x^2+y^2=r^2$

Equation of tangent at circle is

$\Rightarrow$ $x\cos \theta +y\sin \theta =r$

or equation of tangent at ellipse is,

$\Rightarrow$ $y-mx+\sqrt{a^2{m}^2+b^2}$

If it is tangent to circle, then perpendicular from $(0,0)$ is equal to $r.$

$\Rightarrow$ $\frac{\sqrt{a^2{m}^2+b^2}}{\sqrt{m^2+1}}=|r|$

or $a^2{m}^2+b^2=m^2{r}^2+r^2$

or $(a^2-r^2)m^2=r^2-b^2$

$\therefore$ $m=\sqrt{\frac{r^2-b^2}{a^2-r^2}}$

$\therefore$ $\tan \theta =\sqrt{\frac{r^2-b^2}{a^2-r^2}}$

$\therefore$ $\theta ={\tan }^{-1}\sqrt{\frac{r^2-b^2}{a^2-r^2}}$