JEE Main 2024 — 29 January, Shift 2 — Question 23

Parabola is $\mathrm{x}^{2}=8 \mathrm{y}$

Chord with mid point $\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)$ is $\mathrm{T}=\mathrm{S}_{1}$

$\therefore \mathrm{xx}_{1}-4\left(\mathrm{y}{+} \mathrm{y}_{1}\right)=\mathrm{x}_{1}{ }^{2}-8 \mathrm{y}_{1}$

$\therefore\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)=\left(1, \frac{5}{4}\right)$

$\Rightarrow x-4\left(y+\frac{5}{4}\right)=1-8 \times \frac{5}{4}=-9$

$\therefore x-4 y+4=0$

$(\alpha, \beta)$ lies on (i) & also on $y^{2}=4 x$

$\therefore \alpha-4 \beta+4=0$

$\& \beta^{2}=4 \alpha$ Solving (ii) & (iii) $\beta^{2}=4(4 \beta-4) \Rightarrow \beta^{2}-16 \beta+16=0$

$\therefore \beta=8 \pm 4 \sqrt{3}$ and $\alpha=4 \beta-4=28 \pm 16 \sqrt{3}$

$\therefore(\alpha, \quad \beta)=(28+16 \sqrt{3}, 8+4 \sqrt{3})$ and

$(28-16 \sqrt{3}, 8-4 \sqrt{3})$

$\therefore(\alpha-28)(\beta-8)=( \pm 16 \sqrt{3})( \pm 4 \sqrt{3})$

$=192$