Ellipse Question 6
Question 6 - 2024 (31 Jan Shift 2)
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
(1) $\frac{385}{8}$
(2) $\frac{347}{8}$
(3) $\frac{512}{25}$
(4) $\frac{656}{25}$
Show Answer
Answer (4)
Solution
Slope of axis $=\frac{1}{2}$
$y-3=\frac{1}{2}(x-2)$
$\Rightarrow 2 y-6=x-2$
$\Rightarrow 2 y-x-4=0$
$2 x+y-6=0$
$4 x+2 y-12=0$
$\alpha+1.6=4 \Rightarrow \alpha=2.4$
$\beta+2.8=6 \Rightarrow \beta=3.2$
Ellipse passes through $(2.4,3.2)$
$\Rightarrow \frac{\left(\frac{24}{10}\right)^{2}}{a^{2}}+\frac{\left(\frac{32}{10}\right)^{2}}{b^{2}}=1$
Also $1-\frac{b^{2}}{a^{2}}=\frac{1}{2}=\frac{b^{2}}{a^{2}}=\frac{1}{2}$
$\Rightarrow a^{2}=2 b^{2}$
Put in $(1) \Rightarrow b^{2}=\frac{328}{25}$
$\Rightarrow\left(\frac{2 b^{2}}{a}\right)^{2}=\frac{4 b^{2}}{a^{2}} \times b^{2}=4 \times \frac{1}{2} \times \frac{328}{25}=\frac{656}{25}$
