<s>[INST] For an ellipse with minor radius $b$, show that the product of distances from the foci to any tangent line is $b^2$ 
Consider the ellipse with equation:
  $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$
  How do we prove the product of distances from the foci to any tangent line is $b^2$?



$$F_1A \cdot F_2B = b^2$$

 [/INST] Let' say that $A(x_0,y_0)$ is the point of intersection  of the tangent 
$$\epsilon:\begin{array}[t]{l}\frac {x} {a^2} x_0+\frac y {b^2}y_0=1\\(b^2 x_0) x+(a^2 y_0)y -a^2b^2=0\end{array}$$
and the ellipse.
The distance between $F_1(-c,0)$ and $\epsilon$ is:
$$d_1=\dfrac{\left|-c b^2 x_0-a^2b^2\right|}{\sqrt{b^4x_0^2+a^4y_0^2}}$$
Respectively, the distance between $F_2(c,0)$ and $\epsilon$ is:
$$d_2=\dfrac{\left|c b^2 x_0-a^2b^2\right|}{\sqrt{b^4x_0^2+a^4y_0^2}}$$
Thus, we have:
$$d_1\cdot d_2\begin{array}[t]{l}=\cdots =\dfrac{\left|a^4b^4-b^4x_0^2c^2\right|}{b^4x_0^2+a^4y_0^2}=\dfrac{\left|a^4b^4-b^4x_0^2(a^2-b^2)\right|}{b^4x_0^2+a^4y_0^2}\\\\
\end{array}$$
However, from the initial equation of the ellipse, since $A(x_0,y_0)$ belongs to the ellipse, we can prove that:
$$a^4b^4=a^2 b^4x_0^2+b^2a^4y_0^2$$
Making the substitution above, we have:
$$d_1\cdot d_2=\dfrac{\left|a^2 b^4x_0^2+b^2a^4y_0^2-b^4a^2x_0^2+b^6x_0^2\right|}{b^4x_0^2+a^4y_0^2}=\cdots=b^2$$
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