JEE Main 2024 — Parabola Question with Solution

Given,

Parabola $x^2=8y$

We know that,

Chord with mid point $\left(x_1,y_1\right)$ is $T=S_1$

$x{x}_1-8\left(\frac{y+y_1}{2}\right)=x_1{}^2-8y_1$

$\Rightarrow x{x}_1-4\left(y+y_1\right)=x_1{}^2-8y_1$

Now, putting the point $\left(x_1,y_1\right)=\left(1,\frac{5}{4}\right)$ we get,

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

$\therefore x-4y+4=0 \dots \left(\mathrm{i}\right)$

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

$\therefore \alpha -4\beta +4=0 \dots \left(\mathrm{ii}\right)$

And ${\beta }^2=4\alpha \dots \left(\mathrm{iii}\right)$

Solving (ii) and (iii)

${\beta }^2=4(4\beta -4)$

$\Rightarrow {\beta }^2-16\beta +16=0$

$\Rightarrow \beta =8\pm 4\sqrt{3}$ 

And $\alpha =4\beta -4=28\pm 16\sqrt{3}$

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

Now, solving

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

$\Rightarrow (\alpha -28)(\beta -8)=192$