JEE Main 2024 — 31 January, Shift 2 — Question 7

Slope of axis $=\frac{1}{2}$

$y-3=\frac{1}{2}(x-2)$

$\Rightarrow 2 \mathrm{y}-6=\mathrm{x}-2$

$\Rightarrow 2 \mathrm{y}-\mathrm{x}-4=0$

$2 \mathrm{x}+\mathrm{y}-6=0$

$4 \mathrm{x}+2 \mathrm{y}-12=0$

$\alpha+1.6=4 \Rightarrow \alpha=2.4$

$\beta+2.8=6 \Rightarrow \beta=3.2$

Ellipse passes through $(2.4,3.2)$

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

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

$\Rightarrow \mathrm{a}^{2}=2 \mathrm{~b}^{2}$

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

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