Let chord of contact be drawn from every point on the circle $x^2+y^2=100$ to the ellipse [CONT..] 
Let chord of contact be drawn from every point on the circle $x^2+y^2=100$ to the ellipse $\frac{x^2}{4}+\frac{y^2}{9}=1$ such that all lines touch a standard ellipse. Find $e$ for the ellipse

Let the point $(h,k)$ lie on the given circle 
The chord of the contact drawn to the given ellipse is 
$$\frac{hx}{4}+\frac{ky}{9}-1=0$$
This line is coincident with the the tangent to the ellipse  $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
$$y=mx\pm \sqrt{a^2m^2+b^2}$$
Then comparing the two equations 
$$m=\frac{-9h}{4k}$$
And $$\frac{81}{k^2}=a^2m^2+b^2$$
$$\frac{81}{k^2}=\frac{81a^2h^2}{16k^2}+b^2$$
$$(81)(16)=81a^2h^2+16k^2b^2$$
How do I proceed from here? Simply substituting $h^2=100-k^2$ doesn’t give any details for $a$ and $b$

Apply the same method that I did in this answer to a previous question of yours.  
Using the usual parameterization of a circle, we get the one-parameter family of polar lines $$\frac52x\cos t+\frac{10}9y\sin t-1=0.$$ Equate coefficients with the generic line $\lambda x+\mu y+\tau=0$ and eliminate $t$ to obtain $$\frac4{25}\lambda^2+\frac{81}{100}\mu^2-\tau^2=0.$$ This conic is dual to $$\frac{25}4x^2+\frac{100}{81}y^2=1.$$
