Consider the parabola P: y^2=4kx and the ellipse E:x^2a^2+y^2b^2=1.

Question

Consider the parabola $P: y^2=4kx$ and the ellipse $E:\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$. Let the line segment joining the points of intersection of $P$ and $E$ be a common latus rectum of both curves. If the eccentricity of $E$ is $e$, then $e^2+2\sqrt2$ is equal to \_\_\_\_\_\_\_\_.

Answer: 3

Accepted Answer

The latus rectum of $P$ is the line $x=k$, with endpoints $(k,\pm2k)$.

Since this chord is vertical, it must be the ellipse's latus rectum too, so the ellipse's major axis is along the $x$-axis, with foci $(\pm ae,0)$, and the latus rectum through $(ae,0)$ has endpoints $\left(ae,\pm\dfrac{b^2}{a}\right)$.

Matching: $k=ae$ and $2k=\dfrac{b^2}{a}\Rightarrow 2ae=\dfrac{b^2}{a}\Rightarrow b^2=2a^2e$.

Also $b^2=a^2(1-e^2)$, so $a^2(1-e^2)=2a^2e \Rightarrow e^2+2e-1=0$.

$$e=\frac{-2+\sqrt{4+4}}{2}=\sqrt2-1.$$

$$e^2=(\sqrt2-1)^2=3-2\sqrt2 \;\Rightarrow\; e^2+2\sqrt2 = 3.$$

Solutions · 1