Question

$P$ is a point on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$.Then prove $PS+P{S}^{\prime }=$ constant. Where $S$ and $S^{\prime }$ be two focii.

Answer 1

Let the co-ordinate of $P$ be $(a\cos \theta ,b\sin \theta )$ on the ellipse, where $\theta$ be the eccentric angle of the point $P$.

Now co-ordinate of $S$ and $S^{\prime }$ be respectively $(ae,0)$ and $(-ae,0)$ where $e=\sqrt{1-\frac{b^2}{a^2}}$.

Now,

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

$=\sqrt{a^2{\cos }^2\theta -2a^{2}e\cos \theta +a^2{e}^2+b^2{\sin }^2\theta }$

$=\sqrt{a^2{\cos }^2\theta -2a^{2}e\cos \theta +(a^2-b^2)+b^2{\sin }^2\theta }$ [Using value of $e$]

$=\sqrt{a^2{\cos }^2\theta -2a^{2}e\cos \theta +a^2-b^2{\cos }^2\theta }$

$=\sqrt{a^2{e}^2{\cos }^2\theta -2a^{2}e\cos \theta +a^2}$

$=\sqrt{(a-ae\cos \theta {)}^2}=a-ae\cos \theta$.

Similarly we will have $P{S}^{\prime }=a+ae\cos \theta$.

$PS+P{S}^{\prime }=2a=$ constant.