<s>[INST] 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 $b$
 [/INST] According to $\frac{hx}{4}+\frac{ky}{9}=1$, the horizontal and vertical lines corresponding to circular points $(h,k) = (0,10),\> (10,0)$ are $ y= \frac9{10}$, $ x = \frac4{10}$. which also corresponds to the elliptical axes $a= \frac4{10}$ and $b= \frac9{10}$. Thus, the equation of the standard ellipse is 
$$\frac{x^2}{(\frac4{10})^2}+\frac{y^2}{(\frac9{10})^2}=1$$
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