JEE MAIN - Mathematics (2024 - 31st January Evening Shift - No. 6)

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
$$\frac{512}{25}$$
$$\frac{656}{25}$$
$$\frac{385}{8}$$
$$\frac{347}{8}$$

Explanation

JEE Main 2024 (Online) 31st January Evening Shift Mathematics - Ellipse Question 13 English Explanation

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

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) $$\Rightarrow b^2=\frac{328}{25}$$

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

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