Let  $P(\alpha ,\beta )$ be a point on the parabola  $y^2=4x$. If P also lies on the chord of the parabola $x^2=8y$ whose mid point is  $(1,\frac{5}{4}),$ then $(\alpha -28)(\beta -8)$  is equal to ______

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

$\begin{array}{r}\therefore {\mathrm{xx}}_1-4\left({\mathrm{y}}^{+}{\mathrm{y}}_1\right)={\mathrm{x}}_1^2-8{\mathrm{y}}_1\\ \therefore \left({\mathrm{x}}_1,{\mathrm{y}}_1\right)=\left(1,\frac{5}{4}\right)\\ \Rightarrow x-4\left(y+\frac{5}{4}\right)=1-8\times \frac{5}{4}=-9\\ \therefore \mathrm{x}-4\mathrm{y}+4=0\dots \left(i\right)\end{array}$

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

$\begin{array}{r}\therefore \alpha -4\beta +4=0\dots \left(ii\right)\\ \mathrm{\&}{\beta }^2=4\alpha \dots \left(iii\right)\end{array}$

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