Question

The product of the perpendiculars drawn from the two foci of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a>b)$ to the tangent at any point of the ellipse is:

• A. $a^2$
• B. $4a^2$
• C. $4b^2$
• D. $b^2$

Answer 1

Answer: (D)

$b^2$

The equation of the tangent at any point $(a\cos \theta ,b\sin \theta )$ on the ellipse is

$\frac{x}{a}\cos \theta +\frac{y}{b}\sin \theta =1i.e.,bx\cos \theta +ay\sin \theta -ab=0.$

The product of the perpendiculars from $S(ae,0)$ and $S^{{}^{\prime }}(-ae,0)$ on this tangent

$=\frac{(abe\cos \theta -ab)(-abe\cos \theta -ab)}{b^2{\cos }^2\theta +a^2{\sin }^2\theta }$

$=\frac{a^2{b}^2(1-e^2{\cos }^2\theta )}{a^2(1-e^2){\cos }^2\theta +a^2{\sin }^2\theta }=\frac{a^2{b}^2(1-e^2{\cos }^2\theta )}{a^2(1-e^2{\cos }^2\theta )}=b^2$

Ans: D