JEE 2024 — Mathematics PYQ

$\begin{array}{rl}& amp;{\text{ eq }}^n\text{ of chord bisected at point }\left(1,\frac{5}{4}\right)\text{ to Parabola }\\ & amp;x^2=8y\\ & amp;T=S_1\\ & amp;x{x}_1-4\left(y+y_1\right)=x_1^2-8y_1\\ & amp;\therefore \left(x_1,y_1\right)=\left(1,\frac{5}{4}\right)\\ & amp;\Rightarrow x-4y-5=-9\\ & amp;\Rightarrow (\alpha ,\alpha )\text{ lies on }\\ & amp;x-4y+4=0\\ & amp;\&y^2=4x\\ & amp;{\alpha }^2-4\beta +4=0\\ & amp;{\alpha }^2=4\alpha \\ & amp;\Rightarrow {\beta }^1=4(4\alpha -4)\end{array}$

$\Rightarrow {\beta }^1-16\alpha +16=0$

$\Rightarrow \beta =\frac{16\pm \sqrt{(16{)}^2-4(1)(16)}}{2}$

$\beta =8\pm 4\sqrt{3}$

Now

$\alpha =4\beta -4$

$8-\beta =\pm 4\sqrt{3}$

$\alpha =(\pm 16\sqrt{3})\left(\mp 4\sqrt{3}\right)$

$\therefore (\alpha -28)(8-\beta )=\left(\pm 16\sqrt{3}\right)\left(\mp 4\sqrt{3}\right)=-192$