Ellipse 2 Question 15
15. If $a>2 b>0$, then positive value of $m$ for which $y=m x-b \sqrt{1+m^{2}}$ is a common tangent to $x^{2}+y^{2}=b^{2}$ and $(x-a)^{2}+y^{2}=b^{2}$ is
$(2002,1$ M)
(a) $\frac{2 b}{\sqrt{a^{2}-4 b^{2}}}$
(b) $\frac{\sqrt{a^{2}-4 b^{2}}}{2 b}$
(c) $\frac{2 b}{a-2 b}$
(d) $\frac{b}{a-2 b}$
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Answer:
Correct Answer: 15. (a)
Solution:
- Given, $y=m x-b \sqrt{1+m^{2}}$ touches both the circles, so distance from centre $=$ radius of both the circles.
$$ \begin{aligned} & \frac{\left|m a-0-b \sqrt{1+m^{2}}\right|}{\sqrt{m^{2}+1}}=b \text { and } \frac{\left|-b \sqrt{1+m^{2}}\right|}{\sqrt{m^{2}+1}}=b \\ \Rightarrow & \left|m a-b \sqrt{1+m^{2}}\right|=\left|-b \sqrt{1+m^{2}}\right| \\ \Rightarrow & m^{2} a^{2}-2 a b m \sqrt{1+m^{2}}+b^{2}\left(1+m^{2}\right)=b^{2}\left(1+m^{2}\right) \\ \Rightarrow & m a-2 b \sqrt{1+m^{2}}=0 \\ \Rightarrow & m^{2} a^{2}=4 b^{2}\left(1+m^{2}\right) \\ \therefore & m=\frac{2 b}{\sqrt{a^{2}-4 b^{2}}} \end{aligned} $$
