Question

If a variable point P on an ellipse of eccentricity e is joined to the foci $S_1$ and $S_2$ then the incentre of the triangle $PS_1S_2$ lies on

• A. The major axis of the ellipse
• B. The circle with radius e
• C. Another ellipse of eccentricity $\sqrt{\frac{3+e^{2}}{4}}$
• D. None of these

Answer 1

Answer: (C)

Another ellipse of eccentricity $\sqrt{\frac{3+e^{2}}{4}}$

Answer 2

Let the ellipse be $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1\quad....\left(1\right)$ Then $e^{2} = 1-\frac{b^{2}}{a^{2}}\quad ....\left(2\right)$ Let a point P on $\left(1\right)$ be $\left(a\, cos\theta,\, b\, sin\theta\right)$. The coordinates of foci are $S_{1} \left(ae, 0\right)$ and $S_{2} \left(-ae, \,0\right)$. Henc $S_{1}P = a\left(1 - ecos \,\theta \right)$ $S_{2}P = a\left(1 + ecos \,\theta \right)$ and $S_{1}S_{2} = 2ae$ If $\left(h, \,k\right)$ be the coordinates of in centre then $h = \frac{2ae\times \,a \,cos \,\theta + a\left(1 - e \,cos\, \theta \right)\times-ae +a\left(1 + e\, cos\, \theta \right)\times ae}{2ae+a\left(1 - e\, cos\, \theta \right)+a\left(1 + e\, cos \,\theta \right)}$ $= \frac{2ae \,cos\, \theta}{1+e}\quad ....\left(3\right)$ $k = \frac{be\, sin\, \theta}{1+e} \quad ....\left(4\right)$ Squaring and adding $\left(3\right) \,\&\, \left(4\right)$ we have, $\frac{h^{2}}{4a^{2}} + \frac{k^{2}}{b^{2}} = \left(\frac{e}{1+e}\right)^{2}$ $\therefore$ The locus of the point $\left(h, k\right)$ is $\frac{x^{2}}{4a^{2}\lambda^{2}}+\frac{y^{2}}{b^{2}\lambda^{2}} = 1$, where $\lambda = \frac{e}{1+e}$. Which is another ellipse with eccentricity $= \sqrt{1-\frac{b^{2}}{4a^{2}} } = \sqrt{\frac{3+e^{2}}{4}}$