JEE Main 2024 — Parabola Question with Solution

Parabola is $$x^2=8y$$

Chord with mid point $$\left({\mathrm{x}}_1,{\mathrm{y}}_1\right)$$ is $$\mathrm{T}={\mathrm{S}}_1$$

$$\begin{array}{rl}& \therefore {\mathrm{xx}}_1-4\left(\mathrm{y}+{\mathrm{y}}_1\right)={\mathrm{x}}_1^2-8{\mathrm{y}}_1\\ & \therefore \left({\mathrm{x}}_1,{\mathrm{y}}_1\right)=\left(1,\frac{5}{4}\right)\\ & \Rightarrow \mathrm{x}-4\left(\mathrm{y}+\frac{5}{4}\right)=1-8\times \frac{5}{4}=-9\end{array}$$

$$\therefore x-4y+4=0$$ ...... (i)

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

$$\begin{array}{rl}& \therefore \alpha -4\beta +4=0\text{ .... (ii)}\\ & \&{\beta }^2=4\alpha \text{ .... (iii)}\end{array}$$

Solving (ii) & (iii)

$$\begin{array}{rl}& {\beta }^2=4(4\beta -4)\Rightarrow {\beta }^2-16\beta +16=0\\ & \therefore \beta =8\pm 4\sqrt{3}\text{ and }\alpha =4\beta -4=28\pm 16\sqrt{3}\\ & \therefore (\alpha ,\beta )=(28+16\sqrt{3},8+4\sqrt{3})\&\\ & (28-16\sqrt{3},8-4\sqrt{3})\\ & \therefore (\alpha -28)(\beta -8)=(\pm 16\sqrt{3})(\pm 4\sqrt{3})\\ & =192\end{array}$$