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, $S P: S^{\prime} P=$

• 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 $S(a e, 0)$ 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 $S^{\prime}(-a e, 0)$ 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, $S P=a-a e \cos \alpha$
and $S^{\prime} P=a+a e \cos \alpha$
or $\frac{S P}{S^{\prime} P}=\frac{1-e \cos \alpha}{1+e \cos \alpha}=\frac{d_{1}}{d_{2}}$