Question

Find the focal distance of a point $P\left(\theta \right)$ on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a>b)$

Answer 1

Let 'e' be the eccentricity of ellipse $\therefore$ PS = e . PM = e $(\frac{a}{e}-a\cos \theta )PS=(a-ae\cos \theta )$
and PS' = e PM' = e $(a\cos \theta +\frac{a}{e})P{S}^{\prime }=a+ae\cos \theta$
$\therefore$ focal distance are (a $\pm$ ae $\cos \theta )$
Note : PS + PS' = 2a
PS + PS' = AA'