Question

A circle of radius $r(<a)$ is concentric with ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,(a>b)$, then slope of the common tangents to ellipse and circle is

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

Answer 1

The correct option is D $\pm \sqrt{\frac{r^2-b^2}{a^2-r^2}}$

Slope of the tangent to the ellipse
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is
$y=mx\pm \sqrt{a^2{m}^2+b^2}\cdots \left(1\right)$
$\because$ equation $\left(1\right)$ tangent to circle with centre origin (concentric condition) and radius $r$
$\therefore r=\frac{|\pm \sqrt{a^2{m}^2+b^2}|}{\sqrt{1+m^2}}\Rightarrow r^2=\frac{a^2{m}^2+b^2}{1+m^2}\Rightarrow m^2(a^2-r^2)=r^2-b^2\Rightarrow m=\pm \sqrt{\frac{r^2-b^2}{a^2-r^2}}$