Question

Let chord of contact is drawn from every point lying on circle $x^2 + y^2 = 100$ to the ellipse $\frac{x^2}{4} + \frac{y^2}{9} = 1$ such that all the lines touches a standard ellipse whose eccentricity is $e$, then $\frac{9e^2}{13}$ is

Answer 1

$\displaystyle \frac{5}{9}$

Answer 2

1. Equation of chord of contact.
From point $(h,k)$ to ellipse $\frac{x^2}{4} + \frac{y^2}{9}=1$:

$$\frac{h\,x}{4} + \frac{k\,y}{9} = 1.$$

2. Envelope condition.
This family of lines is tangent to the circle $x^2 + y^2 = 100$.
Compare $\frac{h}{4}x + \frac{k}{9}y =1$ with $y=mx+c$:

$$m = -\frac{h/4}{k/9} = -\frac{9h}{4k},\quad c = \frac{1}{k/9} = \frac{9}{k}.$$

Tangent to circle $x^2+y^2=10^2$ satisfies $10 = \frac{|c|}{\sqrt{1+m^2}}$.
Squaring gives

$$100 = \frac{c^2}{1+m^2} = \frac{\bigl(\tfrac{9}{k}\bigr)^2}{1 + \bigl(\tfrac{9h}{4k}\bigr)^2} = \frac{81/k^2}{1 + \frac{81h^2}{16k^2}} = \frac{81}{k^2 + \tfrac{81}{16}h^2}.$$

Thus

$$k^2 + \frac{81}{16}h^2 = \frac{81}{100} \;\Longrightarrow\; \frac{h^2}{\tfrac{16}{100}} + \frac{k^2}{\tfrac{81}{100}} = 1.$$

This is an ellipse with semi‑axes $a=\tfrac{4}{10},\,b=\tfrac{9}{10}$.
Its eccentricity

$$e = \sqrt{1 - \frac{a^2}{b^2}} = \sqrt{1 - \frac{(4/10)^2}{(9/10)^2}} = \sqrt{1 - \frac{16}{81}} = \frac{\sqrt{65}}{9}.$$

3. Compute $\tfrac{9e^2}{13}$.

$$e^2 = \frac{65}{81} \;\Longrightarrow\; \frac{9e^2}{13} = \frac{9\cdot\frac{65}{81}}{13} = \frac{65}{117} = \frac{5}{9}.$$