Analytic Geometry – Problem 14: find the eccentricity of

For ellipse $E:\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\,(a>b>0)$, let $F_1,F_2$ be the left and right foci. Let $P$ be any point on $E$ not coinciding with a vertex, and let $I$ be the incenter of $\triangle PF_1F_2$. Let $O$ be the origin. Denote the slopes of lines $OP$ and $OI$ by $k_1$ and $k_2$, respectively. If $k_1=\dfrac32k_2$, find the eccentricity of $E$.

Let $P(x_P,y_P)$ be on the ellipse, $I(x_I,y_I)$ be the incenter of $\triangle PF_1F_2$, and $c$ be focal half-distance. From the tangent-length/bisector ratio relation in this triangle, $$x_I=e\,x_P,\qquad y_I=\frac{c}{a+c}y_P.$$ Hence $$k_1=\frac{y_P}{x_P},\qquad k_2=\frac{y_I}{x_I}=\frac{\frac{c}{a+c}y_P}{e x_P}=\frac{c}{a+c}\cdot\frac{1}{e}k_1.$$ Since $e=\dfrac{c}{a}$, $$k_2=\frac{a}{a+c}k_1.$$ Given $k_1=\dfrac32k_2$, $$1=\frac32\cdot\frac{a}{a+c}\Rightarrow 2(a+c)=3a\Rightarrow c=\frac{a}{2}.$$ Therefore $$e=\frac{c}{a}=\frac12.$$

$$e=\frac12.$$