If S and $mathrm{S}^{prime}$ are the foci of an ellipse and P any point on the curve, $tan left(frac{mathrm{PSS}^{prime}}{2}right) tan left(frac{mathrm{PS}^{prime} mathrm{S}^{prime}}{2}right)=frac{1-mathrm{e}}{1+mathrm{e}}$. The locus of the center of the circle inscribed in the triangle $mathrm{PSS}^{prime}$ is :

If <img alt="\mathrm{\mathrm{S}\, \, and\, \, \mathrm{S}^{\prime}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Cmathrm%7BS%7D%5C%2C%20%5C%2C%20and%5C%2C%20%5C%2C%20%5C

If $\mathrm{\mathrm{S}\, \, and\, \, \mathrm{S}^{\prime}}$ are the foci of an ellipse and $\mathrm{ \mathrm{P}}$ any point on the curve, $\mathrm{ \tan \left(\frac{\text { PSS }^{\prime}}{2}\right) \tan \left(\frac{\text { PS'S }^{\prime}}{2}\right)=\frac{1-\mathrm{e}}{1+\mathrm{e}}}$ . The locus of the center of the circle inscribed in the triangle $\mathrm{ PSS^{\prime}}$ is :
Option: 1
Straight line

Option: 2
Ellipse

Option: 3
Parabola

Option: 4
Circle

Answers (1)

Let $\mathrm{\angle \mathrm{PSS}^{\prime}=\alpha and \angle \mathrm{PS}^{\prime} \mathrm{S}=\beta }$
$\mathrm{\therefore \quad \mathrm{SS}^{\prime}=2 \mathrm{ae} }$
By sine rule in $\mathrm{ \triangle \mathrm{PSS}^{\prime} }$ then
$\mathrm{\begin{array}{ll} & \frac{\mathrm{SP}}{\sin \beta}=\frac{\mathrm{S}^{\prime} \mathrm{P}}{\sin \alpha}=\frac{\mathrm{SS}^{\prime}}{\sin \{\pi-(\alpha+\beta)\}} \\ \Rightarrow \quad & \frac{\mathrm{SP}}{\sin \beta}=\frac{\mathrm{S}^{\prime} \mathrm{P}}{\sin \alpha}=\frac{\mathrm{SS}^{\prime}}{\sin (\alpha+\beta)} \\ \text { or } & \frac{\mathrm{SP}+\mathrm{S}^{\prime} \mathrm{P}}{\sin \beta+\sin \alpha}=\frac{\mathrm{SS}^{\prime}}{\sin (\alpha+\beta)} \\ \text { or } & \frac{2 \mathrm{a}}{\sin \beta+\sin \alpha}=\frac{2 \mathrm{ae}}{\sin (\alpha+\beta)} \end{array} }$

$\mathrm{\begin{array}{ll} \Rightarrow \quad \frac{1}{\mathrm{e}}=\frac{2 \sin \left(\frac{\alpha+\beta}{2}\right) \cos \left(\frac{\alpha-\beta}{2}\right)}{2 \sin \left(\frac{\alpha+\beta}{2}\right) \cos \left(\frac{\alpha+\beta}{2}\right)} \\ \Rightarrow \quad \frac{1}{\mathrm{e}}=\frac{\cos \left(\frac{\alpha-\beta}{2}\right)}{\cos \left(\frac{\alpha+\beta}{2}\right)} \\ \Rightarrow \quad \frac{1-\mathrm{e}}{1+\mathrm{e}}=\frac{\cos \left(\frac{\alpha-\beta}{2}\right)-\cos \left(\frac{\alpha+\beta}{2}\right)}{\cos \left(\frac{\alpha-\beta}{2}\right)+\cos \left(\frac{\alpha+\beta}{2}\right)} \\ \Rightarrow \quad \frac{1-\mathrm{e}}{1+\mathrm{e}}=\tan \alpha / 2 \tan \beta / 2 \\ \text { or } \quad \tan \left(\frac{\mathrm{PSS}^{\prime}}{2}\right) \tan \left(\frac{\mathrm{PS} \mathrm{S}^{\prime}}{2}\right)=\frac{1-\mathrm{e}}{1+\mathrm{e}} \end{array} }$ .....................(1)
The centre of the circle inscribed in $\mathrm{\triangle \mathrm{PSS}^{\prime} }$ will lie on the bisectors of angles $\mathrm{\alpha }$ and $\mathrm{\beta }$ whose equations are
$\mathrm{\begin{aligned} & y=\tan \frac{\beta}{2}(x+a e) \\ & y=\tan \left(\pi-\frac{\alpha}{2}\right)(x-a e) \end{aligned} }$
and
$\mathrm{y=-\tan \frac{\alpha}{2}(x-a e) }$
or
By multiplying we get
$\mathrm{\begin{aligned} & \mathrm{y}^2=-\tan \frac{\alpha}{2} \tan \frac{\beta}{2}\left(\mathrm{x}^2-\mathrm{a}^2 \mathrm{e}^2\right) \\ & \mathrm{y}^2=-\frac{1-\mathrm{e}}{1+\mathrm{e}}\left(\mathrm{x}^2-\mathrm{a}^2 \mathrm{e}^2\right) \\ & \mathrm{y}^2+\mathrm{x}^2\left(\frac{1-\mathrm{e}}{1+\mathrm{e}}\right)=\mathrm{a}^2 \mathrm{e}^2\left(\frac{1-\mathrm{e}}{1+\mathrm{e}}\right) \\ & \frac{\mathrm{x}^2}{\mathrm{a}^2 \mathrm{e}^2}+\frac{\mathrm{y}^2}{\mathrm{a}^2 \mathrm{e}^2\left(\frac{1-\mathrm{e}}{1+\mathrm{e}}\right)}= \end{aligned} }$ (From equation (1))
or
or which is an ellipse.