Question

A straight line drawn through the common focus S' of a number of conics meets them in the points $P_1,P_2,....;$ on it is taken a point Q such that the reciprocal of SQ is equal to the sum of the reciprocals of $S{P}_1,S{P}_2,...$ Prove that the locus of Q is a conic section whose focus is O, and show that the reciprocal of its latus rectum is equal to the sum of the reciprocals of the latera recta of the given conics.

Answer 1

Equation of a conic in polar form is

$\frac{l}{r}=1-e\cos \theta$

Let the eccentricities of the conies be $e_1,e_2,e_3....$ and latustrecum $l_1,l_2,l_3....$ . Let the axes be inclined at angle ${\alpha }_1,{\alpha }_2,{\alpha }_3....$ then

$\frac{l}{{SP}_1}=1-e_1\cos (\theta -{\alpha }_1)\Rightarrow \frac{1}{{SP}_1}=\frac{1-e_1\cos (\theta -{\alpha }_1)}{l}$

Similarly $\frac{l}{{SP}_2}=1-e_2\cos (\theta -{\alpha }_2)$

$\Rightarrow \frac{1}{{SP}_2}=\frac{1-e_2\cos (\theta -{\alpha }_2)}{l}$

Given $\frac{1}{SQ}=\frac{1}{{SP}_1}+\frac{1}{{SP}_2}......$

$\Rightarrow \frac{1}{SQ}=\frac{1-e_1\cos (\theta -{\alpha }_1)}{l_1}+\frac{1-e_2\cos (\theta -{\alpha }_2)}{l_2}......\Rightarrow \frac{1}{SQ}=(\frac{1}{l_1}+\frac{1}{l_2}.....)-(\frac{e_1}{l_1}\cos (\theta -{\alpha }_1)+\frac{e_2}{l_2}\cos (\theta -{\alpha }_2)......)$

This is equation of form $\frac{1}{r}=\frac{1}{l}-\frac{e}{l}\cos (\theta -\alpha )$

Hence it represents a conic with focus $S^{\prime }$ and latusrectum $L$

where $\frac{1}{L}=\frac{1}{l_1}+\frac{1}{l_2}+\frac{1}{l_3}.....$

Hence proved