Question
The locus of incentre of $\Delta PF_{1}F_{2}$ where $P$ is a variable point lying on ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ with $F_{1},F_{2}$ as foci and eccentricity $e$ is a conic whose:
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Solution
Sol. $h = \frac{a^{2}e(1 + ecos\theta) - a^{2}e(1 - ecos\theta)}{+ 2a^{2}ecos\theta}$
$$\Rightarrow h = aecos\theta$$
$$k = \frac{2aebsin\theta}{2a + 2ae} = \frac{ebsin\theta}{(1 + e)}$$
$$\therefore\ \frac{h^{2}}{(ae)^{2}} + \frac{k^{2}}{\left( \frac{be}{1 + e} \right)^{2}} = 1$$
Equation of locus of incentre is
$$\begin{matrix} & \frac{x^{2}}{(ae)^{2}} + \frac{y^{2}}{\left( \frac{be}{1 + e} \right)^{2}} = 1 \\ \therefore & \ \frac{b^{2}e^{2}}{(1 + e)^{2}} = a^{2}e^{2}\left( 1 - e^{'} \right)^{2} \\ \Rightarrow & \frac{a^{2}e^{2}\left( a - e^{2} \right)}{(1 + e)^{2}} = a^{2}e^{2}\left( 1 - e^{'} \right)^{2} \\ \Rightarrow & e^{'2} = 1 - \frac{1 - e}{1 + e} = \frac{2e}{1 + e} \\ \Rightarrow & e^{'} = \sqrt{\frac{2e}{1 + e}} \end{matrix}$$
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