Ellipse
Ellipse Through Focus of Parabola — Latus Rectum
DAILY_CHALLENGE
Grade 11
Question:
Let $P$ be a parabola with vertex $(2,3)$ and directrix $2x+y=6$. Let an ellipse $E:\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$, $a>b$ of eccentricity $\dfrac{1}{\sqrt{2}}$ pass through the focus of the parabola $P$. Then the square of the length of the latus rectum of $E$ is
$\dfrac{385}{8}$
$\dfrac{347}{8}$
$\dfrac{512}{25}$
$\dfrac{656}{25}$
Step-by-Step Solution
Key Concept: Find the focus of the parabola: reflect vertex $(2,3)$ across the directrix $2x+y=6$. The ellipse has $e=1/\sqrt{2}$ so $b^2=a^2/2$. Use the focus of the parabola as a point on the ellipse to find $a^2$ and $b^2$. Compute $(2b^2/a)^2$.
Focus of parabola $P$: vertex $(2,3)$, directrix $2x+y=6$. Foot of perp: $(1.6,2.8)$. Focus: $(2.4,3.2)$. Ellipse: $e=1/\sqrt2\Rightarrow a^2=2b^2$. Substituting $(2.4,3.2)$: $b^2=328/25$. $(LR)^2=\left(\frac{2b^2}{a}\right)^2=\frac{4b^4}{a^2}=\frac{4b^4}{2b^2}=2b^2=\frac{656}{25}$.
Correct Answer: 4