A variable point P on the ellipse of eccentricity e is joined to the foci S and S'. The eccentricity of the locus of incentre the triangle PSS' is

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$\sqrt{\frac{2 e}{1 + e}}$

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

$\sqrt{\frac{1 - e}{1 + e}}$

$\frac{e}{2 (1 + e)}$

answer is A.

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Detailed Solution

$Incentre = (\frac{2 a^{2} e \cos θ - a^{2} e (1 - e \cos θ) + a^{2} e (1 + e \cos θ)}{2 a e + 2 α}, \frac{2 a b e \sin θ}{2 a (1 + e)})$

$(x, y) = (\frac{2 a^{2} e \cos θ + 2 a^{2} e^{2} \cos θ}{2 a (1 + e)}, \frac{b e \sin θ}{1 + e}) \Rightarrow x = a e \cos θ, y = \frac{b e \sin θ}{1 + e} \Rightarrow \frac{x^{2}}{{(a e)}^{2}} + \frac{y^{2}}{{(\frac{b e}{(1 + e})}^{2}} = 1 \Rightarrow \frac{x^{2}}{{(a e)}^{2}} + \frac{y^{2} {(1 + e)}^{2}}{e^{2} a^{2} (1 - e^{2})} = 1 \Rightarrow \frac{x^{2}}{{(a e)}^{2}} + \frac{y^{2}}{a^{2} e^{2} (\frac{1 - e}{1 + e})} = 1 Eccentricity = \sqrt{1 - \frac{1 - e}{1 + e}} = \sqrt{\frac{2 e}{1 + e}}$