Stephanie Hunter

2022-07-17

Prove that the locus of the incenter of the $\mathrm{\Delta }PS{S}^{\prime }$ 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 $\mathrm{\Delta }PS{S}^{\prime }$ is an ellipse of eccentricity $\sqrt{\frac{2e}{1+e}}$.
Let P be $\left(a\mathrm{cos}\theta ,b\mathrm{sin}\theta \right)$. 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
$\left(\frac{a{x}_{1}+b{x}_{2}+c{x}_{3}}{a+b+c},\frac{a{y}_{1}+b{y}_{2}+c{y}_{3}}{a+b+c}\right)\phantom{\rule{thinmathspace}{0ex}}.$
Then,
but I could not find the relationship between h and k, whence I could not find the eccentricity of this ellipse.

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minotaurafe

Expert

Step 1
Consider an ellipse with semimajor axis a and semiminor axis b centered at (0,0), where $a\ge b>0$. The eccentricity e of this ellipse is given by $e=\sqrt{1-\frac{{b}^{2}}{{a}^{2}}}$. Without loss of generality, let $S=\left(+c,0\right)$ and ${S}^{\prime }=\left(-c,0\right)$ be the foci of this ellipse, where $c:=ae$. Note that $PS+P{S}^{\prime }=2a$ for every point P on the ellipse. Thus, if $\left(h\left(\theta \right),k\left(\theta \right)\right)$ is the incenter of $P=\left(a\phantom{\rule{thinmathspace}{0ex}}\mathrm{cos}\left(\theta \right),b\phantom{\rule{thinmathspace}{0ex}}\mathrm{sin}\left(\theta \right)\right)$ for $\theta \in \left[0,2\pi \right)$, then, as you have found out,
$h\left(\theta \right)=\frac{2{a}^{2}e\phantom{\rule{thinmathspace}{0ex}}\mathrm{cos}\left(\theta \right)-ae\phantom{\rule{thinmathspace}{0ex}}PS+ae\phantom{\rule{thinmathspace}{0ex}}P{S}^{\prime }}{2ae+2a}=\frac{e}{e+1}\left(a\mathrm{cos}\left(\theta \right)-\frac{1}{2}PS+\frac{1}{2}P{S}^{\prime }\right)\phantom{\rule{thinmathspace}{0ex}}.$.
and $k\left(\theta \right)=\frac{2abe\phantom{\rule{thinmathspace}{0ex}}\mathrm{sin}\left(\theta \right)}{2ae+2a}=\frac{e}{e+1}\left(b\phantom{\rule{thinmathspace}{0ex}}\mathrm{sin}\left(\theta \right)\right)\phantom{\rule{thinmathspace}{0ex}}.$.
Step 2
Using $e=\sqrt{1-\frac{{b}^{2}}{{a}^{2}}}$ with
it can be easily seen that $PS=a\left(1-e\phantom{\rule{thinmathspace}{0ex}}\mathrm{cos}\left(\theta \right)\right)$ and $P{S}^{\prime }=a\left(1+e\phantom{\rule{thinmathspace}{0ex}}\mathrm{cos}\left(\theta \right)\right)$. Therefore,
$h\left(\theta \right)=\frac{e}{e+1}a\left(1+e\right)\phantom{\rule{thinmathspace}{0ex}}\mathrm{cos}\left(\theta \right)=ae\phantom{\rule{thinmathspace}{0ex}}\mathrm{cos}\left(\theta \right)\phantom{\rule{thinmathspace}{0ex}}.$
Let $A:=ae$ and $B:=\frac{e}{e+1}b$. Then,

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 $E=\sqrt{1-\frac{{B}^{2}}{{A}^{2}}}=\sqrt{1-\frac{{b}^{2}/{a}^{2}}{\left(e+1{\right)}^{2}}}=\sqrt{1-\frac{1-{e}^{2}}{\left(e+1{\right)}^{2}}}=\sqrt{1-\frac{1-e}{e+1}}=\sqrt{\frac{2e}{e+1}}\phantom{\rule{thinmathspace}{0ex}}.$

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