Let x=9 be a directrix of an ellipse E, whose centre is at the origin and eccentricity is …

Question

Let $x=9$ be a directrix of an ellipse $E$, whose centre is at the origin and eccentricity is $\dfrac13$. Let $P$, with $\alpha>0$, be a focus of $E$, and $AB$ be a chord passing through $P$. Then the locus of the midpoint of $AB$ is:

Accepted Answer

Directrix $x=a/e=9$ with $e=\dfrac13 \Rightarrow a=3,\ c=ae=1$.

$$b^2=a^2(1-e^2)=9\left(1-\frac19\right)=8 \;\Rightarrow\; E:\frac{x^2}{9}+\frac{y^2}{8}=1.$$

Focus $P=(1,0)$. For a chord with midpoint $(h,k)$, the chord equation (T = S\_1) is

$$\frac{xh}{9}+\frac{yk}{8}=\frac{h^2}{9}+\frac{k^2}{8}.$$

Since this chord passes through $P(1,0)$:

$$\frac{h}{9}=\frac{h^2}{9}+\frac{k^2}{8} \;\Rightarrow\; 8h=8h^2+9k^2 \;\Rightarrow\; 9k^2=8h(1-h).$$

Replacing $(h,k)\to(x,y)$: $\;9y^2=8x(1-x)$.

Solutions · 1