Ellipse: product of the distance from foci to a tangent is a constant I am supposed to determine what is the result of said product. Given $P(x_0,y_0)$, I need to calculate the distance from the foci of an ellipse to the tangent line that passes through $P$, and then multiply the distances.
In essence it is quite simply. We take:
$$
\frac{x_0}{a^2}x + \frac{y_0}{b^2}y = 1
$$
as the tangent line. Then we simply calculate its distance to each focus $(c,0)$ and $(-c,0)$, using the formula and then, multiplying.
$$
d=\frac{\frac{x_0c}{a^2}±1}{\sqrt{\frac{x_0^2}{a^4} + \frac{y_0^2}{b^4}}}
$$
$$
\text{Some constant k}=\frac{\frac{x_0^2c^2}{a^4}-1}{\frac{x_0^2}{a^4} + \frac{y_0^2}{b^4}}
$$
I'm having trouble getting things cancelled here. The constant k is $b^2$, but I can't get to it. Help?
 A: Here I go:
$$
\frac   {\frac{x_0^2c^2}{a^4}-1}     {\frac{x_0^2}{a^4} + \frac{y_0^2}{b^4}}
$$
Simplify
$$
\frac   {\frac{x_0^2c^2-a^4}{a^4}}    {\frac{x_0^2b^4 + y_0^2a^4}{a^4b^4}}
$$
Then
$$
\frac{(x_0^2c^2-a^4)(a^4b^4)}{a^4(x_0^2b^4 + y_0^2a^4)} \\
$$
Quick cancellation
$$
\frac{(x_0^2c^2-a^4)(b^4)}{x_0^2b^4 + y_0^2a^4}
$$
We have from the original ellipse equation that $y_0^2a^4 = a^4b^2-a^2b^2x_0^2$, so:
$$
\frac{(x_0^2c^2-a^4)(b^4)}{x_0^2b^4 + a^4b^2-a^2b^2x_0^2}
$$
We factor out $b^4$ above, and $b^2$ below
$$
b^2\frac{x_0^2c^2-a^4}{x_0^2b^2 + a^4-a^2x_0^2}
$$
Factorize the $x_0$ below
$$
b^2\frac{x_0^2c^2-a^4}{x_0^2(b^2 - a^2) + a^4}
$$
Using the Pythagorean identity
$$
b^2\frac{x_0^2c^2-a^4}{x_0^2(-c^2) + a^4}
$$
Move around
$$
b^2\frac{x_0^2c^2-a^4}{-(x_0^2c^2 - a^4)}
$$
Nice cancellation
$$
k = -b^2
$$
The $-$ sign shouldn't matter, since the distance formula uses absolute values, and in the end, it would be $±b^2$, right?
A: $$
\frac{1-\frac{c^2x_0^2}{a^4}}{\frac{x_0^2}{a^4}+\frac{y_0^2}{b^4}}
=\frac{1-\frac{c^2x_0^2}{a^4}}{\frac{x_0^2}{a^4}+\frac1{b^2}\left(1-\frac{x_0^2}{a^2}\right)}
=\frac{1-\frac{c^2x_0^2}{a^4}}{\frac1{b^2}\left(1-(a^2-b^2)\frac{x_0^2}{a^4}\right)}
=b^2,
$$
where in the final step we used the identity $a^2-b^2=c^2$.
