P is a point on the ellipse frac{x^2}{a^2} + | vedclass

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(A) The foci of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ are $S(ae, 0)$ and $S'(-ae, 0)$.Let $P(a \cos 	heta, b \sin 	heta)$ be a point o

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$\frac{be}{1 + e}$

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(A) The foci of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ are $S(ae, 0)$ and $S'(-ae, 0)$.Let $P(a \cos 	heta, b \sin 	heta)$ be a point on the ellipse.The area of $\Delta PSS'$ is given by $	ext{Area} = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes (2ae) 	imes |b \sin 	heta| = abe |\sin 	heta|$.This area is maximum when $|\sin 	heta| = 1$,i.e.,$	heta = \frac{\pi}{2}$ or $\frac{3\pi}{2}$.Thus,$P$ is $(0, b)$ or $(0, -b)$.For $P(0, b)$,the sides of $\Delta PSS'$ are $SS' = 2ae$,$PS = \sqrt{(ae-0)^2 + (0-b)^2} = \sqrt{a^2e^2 + b^2} = \sqrt{a^2e^2 + a^2(1-e^2)} = a$,and $PS' = a$.The semi-perimeter $s = \frac{2ae + a + a}{2} = ae + a = a(1+e)$.The area $\Delta = abe$.The inradius $r = \frac{\Delta}{s} = \frac{abe}{a(1+e)} = \frac{be}{1+e}$.

$P$ is a point on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. When the area of $\Delta PSS'$ is maximum,the inradius of $\Delta PSS'$ ($S$ and $S'$ are foci) is equal to:

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