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?
Best Answer
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?