Question

If $P(x,y)$ be any point on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a>b)$ where S and S' are foci, then prove that $SP+S^{\prime }P$ is a constant.

Answer 1

$PM=x$ coordinate of $M-x$ coordinate of $P$

$\frac{a}{e}-x_1$

$P{M}^{\prime }=x$ coordinate of $P-x$ coordinate of $M^{\prime }$

$x_1-\left(\frac{-a}{e}\right)=x_1+\frac{a}{e}$

We know by definition of ellipse $\frac{SP}{PM}=e$

$\frac{S^{\prime }P}{P{M}^{\prime }}=e$

So, $SP=ePM=e(\frac{a}{e}-x_1)$

$=a-e{x}_1$

So, $S^{\prime }P=eP{M}^{\prime }=e(x_1+\frac{a}{e})$

$e{x}_1+a$

So, $SP+S^{\prime }P=a-e{x}_1+e{x}_1+a$

$=2a$

$\therefore SP+S^{\prime }P=2a$ is constant for any point on ellipse.