Let P be a parabola with vertex (2,3) and directrix 2 x+y=6.

Question

Let $P$ be a parabola with vertex $(2,3)$ and directrix $2 x+y=6$. Let an ellipse $$E: \frac{x^2}{a^2}+\frac{y^2}{b^2}=1, a>b$$, of eccentricity $$\frac{1}{\sqrt{2}}$$ pass through the focus of the parabola $P$. Then, the square of the length of the latus rectum of $E$, is

Accepted Answer

$$\begin{aligned}
& \text { Slope of axis }=\frac{1}{2} \
& y-3=\frac{1}{2}(x-2) \
& \Rightarrow 2 y-6=x-2 \
& \Rightarrow 2 y-x-4=0 \
& 2 x+y-6=0 \
& 4 x+2 y-12=0 \
& \alpha+1.6=4 \Rightarrow \alpha=2.4 \
& \beta+2.8=6 \Rightarrow \beta=3.2
\end{aligned}$$

Ellipse passes through $(2.4,3.2)$

$$\Rightarrow \frac{\left(\frac{24}{10}\right)^2}{\mathrm{a}^2}+\frac{\left(\frac{32}{10}\right)^2}{\mathrm{~b}^2}=1$$ ..... (1)

Also $$1-\frac{\mathrm{b}^2}{\mathrm{a}^2}=\frac{1}{2}=\frac{\mathrm{b}^2}{\mathrm{a}^2}=\frac{1}{2}$$

$$\Rightarrow a^2=2 b^2$$

Put in (1) $$\Rightarrow b^2=\frac{328}{25}$$

$$\Rightarrow\left(\frac{2 \mathrm{~b}^2}{\mathrm{a}}\right)^2=\frac{4 \mathrm{~b}^2}{\mathrm{a}^2} \times \mathrm{b}^2=4 \times \frac{1}{2} \times \frac{328}{25}=\frac{656}{25}
$$

Solutions · 1