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