Question

A variable point $P$ on the ellipse of eccentricity is joined to the foci $S$ and $s^{\prime }$. The eccentricity of the locus of the in cetre of the triangle $PS{S}^1$ is

• A. $\sqrt{\frac{2e}{1+e}}$
• B. $\sqrt{\frac{e}{1+e}}$
• C. $\sqrt{\frac{1-e}{1+e}}$
• D. $\frac{e}{2(1+e)}$

Answer 1

The correct option is A $\sqrt{\frac{2e}{1+e}}$

In centre of $△P{S}_1{S}_2$ is

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

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

$=(\frac{2a^{2}e\cos \theta +(1+e)}{2a(1+e)},\frac{be\sin \theta }{1+e})=(x,y)$

$\frac{x}{ae}=\cos \theta ,\sin \theta =\frac{(1+e)y}{be}$

$\frac{x^2}{a^2{e}^2}+\frac{(1+e{)}^2{y}^2}{a^2(1-e^2)e^2}=1$

$e^1=\sqrt{1-\frac{a^2{e}^2(1-e)}{a^2{e}^2(1+e)}}=\sqrt{\frac{1+e-1+e}{1+e}}=\sqrt{\frac{2e}{1+e}}$