Question

The product of the perpendicular distance from the foci to any tangent to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

Equation of any tangent is $\frac{x}{a}\cos \theta +\frac{y}{b}\sin \theta =1.$ Foci = $(\pm ae,0)$
Product of perpendicular distance from $(\pm ae,0)$ to the tangent is
$=\frac{(e\cos \theta -1)(-e\cos \theta -1)}{a^{\frac{1}{2}}{\cos }^2\theta +\frac{1}{b^2}{\sin }^2\theta }=\frac{1-e^2{\cos }^2\theta }{a^{\frac{1}{2}}{\cos }^2\theta +\frac{1}{b^2}(1-{\cos }^2\theta )}=\frac{1-\frac{a^2-b^2}{a^2}{\cos }^2\theta }{\frac{1}{b^2}-(\frac{1}{b^2}-\frac{1}{a^2}){\cos }^2\theta }=b^2$