# Prove that the locus of the incenter of the $\Delta PSS'$ is an ellipse of eccentricity $\sqrt{\frac{2e}{1+e}}$

Let $S$ and $S'$ be the foci of an ellipse whose eccentricity is $e$.$P$ is a variable point on the ellipse.Prove that the locus of the incenter of the $\Delta PSS'$ is an ellipse of eccentricity $\sqrt{\frac{2e}{1+e}}$.

Let $P$ be $(a\cos \theta,b \sin \theta)$. Let the incenter of the triangle $PSS'$ be $(h,k)$. The formula for the incenter of a triangle whose side lengths are $a,b,c$ and whose vertices have coordinates $\left(x_1,y_1\right)$, $\left(x_2,y_2\right)$, and $\left(x_3,y_3\right)$ is $$\left(\frac{ax_1+bx_2+cx_3}{a+b+c},\frac{ay_1+by_2+cy_3}{a+b+c}\right)\,.$$
Then, $$h=\frac{2c\cdot a\cos \theta+PS'\cdot c-PS\cdot c}{2c+PS'+PS}\text{ and }k=\frac{2c\cdot a\sin \theta}{2c+PS'+PS}\,,$$ but I could not find the relationship between $h$ and $k$, whence I could not find the eccentricity of this ellipse.

• does it help that $PS+PS'$ is constant? – David Quinn Aug 18 '15 at 10:14

Consider an ellipse with semimajor axis $a$ and semiminor axis $b$ centered at $(0,0)$, where $a\geq b>0$. The eccentricity $e$ of this ellipse is given by $e=\sqrt{1-\frac{b^2}{a^2}}$. Without loss of generality, let $S=(+c,0)$ and $S'=(-c,0)$ be the foci of this ellipse, where $c:=ae$. Note that $PS+PS'=2a$ for every point $P$ on the ellipse. Thus, if $\big(h(\theta),k(\theta)\big)$ is the incenter of $P=\big(a\,\cos(\theta),b\,\sin(\theta)\big)$ for $\theta\in[0,2\pi)$, then, as you have found out, $$h(\theta)=\frac{2a^2e\,\cos(\theta)-ae\,PS+ae\,PS'}{2ae+2a}=\frac{e}{e+1}\Big(a\cos(\theta)-\frac{1}{2}PS+\frac{1}{2}PS'\Big)\,.$$ and $$k(\theta)=\frac{2abe\,\sin(\theta)}{2ae+2a}=\frac{e}{e+1}\big(b\,\sin(\theta)\big)\,.$$ Using $e=\sqrt{1-\frac{b^2}{a^2}}$ with $$PS=\sqrt{\big(a\,\cos(\theta)-ae)^2+b^2\sin^2(\theta)}\;\,\text{ and }PS'=\sqrt{\big(a\,\cos(\theta)+ae)^2+b^2\sin^2(\theta)}\;,$$ it can be easily seen that $PS=a\big(1-e\,\cos(\theta)\big)$ and $PS'=a\big(1+e\,\cos(\theta)\big)$. Therefore, $$h(\theta)=\frac{e}{e+1}a(1+e)\,\cos(\theta)=ae\,\cos(\theta)\,.$$ Let $A:=ae$ and $B:=\frac{e}{e+1}b$. Then, $$h(\theta)=A\,\cos(\theta)\text{ and }k(\theta)=B\,\sin(\theta)\,.$$ Therefore, the locus of the incenter of $PSS'$ is indeed an ellipse with the semimajor axis $A$ and the semiminor axis $B$. If $E$ is its eccentricity, then $$E=\sqrt{1-\frac{B^2}{A^2}}=\sqrt{1-\frac{b^2/a^2}{(e+1)^2}}=\sqrt{1-\frac{1-e^2}{(e+1)^2}}=\sqrt{1-\frac{1-e}{e+1}}=\sqrt{\frac{2e}{e+1}}\,.$$