Stephanie Hunter

Answered

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 $(a\mathrm{cos}\theta ,b\mathrm{sin}\theta )$. 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 $({x}_{1},{y}_{1})$, $({x}_{2},{y}_{2})$, and $({x}_{3},{y}_{3})$ is

$(\frac{a{x}_{1}+b{x}_{2}+c{x}_{3}}{a+b+c},\frac{a{y}_{1}+b{y}_{2}+c{y}_{3}}{a+b+c})\phantom{\rule{thinmathspace}{0ex}}.$

Then, $h=\frac{2c\cdot a\mathrm{cos}\theta +P{S}^{\prime}\cdot c-PS\cdot c}{2c+P{S}^{\prime}+PS}\text{and}k=\frac{2c\cdot a\mathrm{sin}\theta}{2c+P{S}^{\prime}+PS}\phantom{\rule{thinmathspace}{0ex}},$

but I could not find the relationship between h and k, whence I could not find the eccentricity of this ellipse.

Answer & Explanation

minotaurafe

Expert

2022-07-18Added 22 answers

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 $S=(+c,0)$ and ${S}^{\prime}=(-c,0)$ 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 $(}h(\theta ),k(\theta ){\textstyle )$ is the incenter of $P={\textstyle (}a\phantom{\rule{thinmathspace}{0ex}}\mathrm{cos}(\theta ),b\phantom{\rule{thinmathspace}{0ex}}\mathrm{sin}(\theta ){\textstyle )}$ for $\theta \in [0,2\pi )$, then, as you have found out,

$h(\theta )=\frac{2{a}^{2}e\phantom{\rule{thinmathspace}{0ex}}\mathrm{cos}(\theta )-ae\phantom{\rule{thinmathspace}{0ex}}PS+ae\phantom{\rule{thinmathspace}{0ex}}P{S}^{\prime}}{2ae+2a}=\frac{e}{e+1}{\textstyle (}a\mathrm{cos}(\theta )-\frac{1}{2}PS+\frac{1}{2}P{S}^{\prime}{\textstyle )}\phantom{\rule{thinmathspace}{0ex}}.$.

and $k(\theta )=\frac{2abe\phantom{\rule{thinmathspace}{0ex}}\mathrm{sin}(\theta )}{2ae+2a}=\frac{e}{e+1}{\textstyle (}b\phantom{\rule{thinmathspace}{0ex}}\mathrm{sin}(\theta ){\textstyle )}\phantom{\rule{thinmathspace}{0ex}}.$.

Step 2

Using $e=\sqrt{1-\frac{{b}^{2}}{{a}^{2}}}$ with $PS=\sqrt{{\textstyle (}a\phantom{\rule{thinmathspace}{0ex}}\mathrm{cos}(\theta )-ae{)}^{2}+{b}^{2}{\mathrm{sin}}^{2}(\theta )}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{and}P{S}^{\prime}=\sqrt{{\textstyle (}a\phantom{\rule{thinmathspace}{0ex}}\mathrm{cos}(\theta )+ae{)}^{2}+{b}^{2}{\mathrm{sin}}^{2}(\theta )}\phantom{\rule{thickmathspace}{0ex}},$

it can be easily seen that $PS=a{\textstyle (}1-e\phantom{\rule{thinmathspace}{0ex}}\mathrm{cos}(\theta ){\textstyle )}$ and $P{S}^{\prime}=a{\textstyle (}1+e\phantom{\rule{thinmathspace}{0ex}}\mathrm{cos}(\theta ){\textstyle )}$. Therefore,

$h(\theta )=\frac{e}{e+1}a(1+e)\phantom{\rule{thinmathspace}{0ex}}\mathrm{cos}(\theta )=ae\phantom{\rule{thinmathspace}{0ex}}\mathrm{cos}(\theta )\phantom{\rule{thinmathspace}{0ex}}.$

Let $A:=ae$ and $B:=\frac{e}{e+1}b$. Then,

$h(\theta )=A\phantom{\rule{thinmathspace}{0ex}}\mathrm{cos}(\theta )\text{and}k(\theta )=B\phantom{\rule{thinmathspace}{0ex}}\mathrm{sin}(\theta )\phantom{\rule{thinmathspace}{0ex}}.$

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

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