Question

Let $d_1$ and $d_2$ be the lengths of the perpendiculars drawn from the foci $S$ and $S^{\prime }$ of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ to the tangent at any point $P$ on the ellipse. Then, $SP:S^{\prime }P=$

• A. $d_1:d_2$
• B. $d_2:d_1$
• C. $d_1^2:d_2^2$
• D. $\sqrt{d_1}:\sqrt{d_2}$

Answer 1

Answer: (A)

Tangent at $P(a\cos \alpha ,b\sin \alpha )$ is
$\frac{x}{a}\cos \alpha +\frac{y}{b}\sin \alpha =1$....(i)
The distance of focus $S(ae,0)$ from this tangent is
$d_1=\frac{|e\cos \alpha -1|}{\sqrt{\frac{{\cos }^2\alpha }{a^2}+\frac{{\sin }^2\alpha }{b^2}}}$
$=\frac{1-e\cos \alpha }{\sqrt{\frac{{\cos }^2\alpha }{a^2}+\frac{{\sin }^2\alpha }{b^2}}}$
The distance of focus $S^{\prime }(-ae,0)$ from this line is
$d_2=\frac{1+e\cos \alpha }{\sqrt{\frac{{\cos }^2\alpha }{a^2}+\frac{{\sin }^2\alpha }{b^2}}}$
or $\frac{d_1}{d_2}=\frac{1-e\cos \alpha }{1+e\cos \alpha }$
Now, $SP=a-ae\cos \alpha$
and $S^{\prime }P=a+ae\cos \alpha$
or $\frac{SP}{S^{\prime }P}=\frac{1-e\cos \alpha }{1+e\cos \alpha }=\frac{d_1}{d_2}$