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 $b^2$?



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


WLONG, we can represent any point on the given Ellipse as $x=a\cos A,y=b\sin A$
So, the equation of the tangent at $(a\cos A,b\sin A)$  will be $\dfrac{x\cdot a\cos A}{a^2}+\dfrac{y\cdot b\sin A}{b^2}=1$
$$\iff bx\cos A+ay\sin A-ab=0$$
The coordinates of the foci are $(\pm ae,0)$ where $e$ is the eccentricity
So, the distances will be $$\frac{|\pm ae\cdot b\cos A-ab|}{\sqrt{(b\cos A)^2+(a\sin A)^2}}$$ and consequently their product will be
$$=a^2b^2\cdot\frac{(1-e\cos A)(1+e\cos A)}{(b\cos A)^2+(a\sin A)^2} =a^2b^2\cdot\frac{(1-e^2\cos^2A)}{a^2(1-e^2)\cos^2A+a^2\sin^2A}=b^2$$
as $b^2=(1-e^2)a^2$
