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

<p>Parabola is $x^2=8 y$</p> <p>Chord with mid point $\left(\mathrm{x}_1, \mathrm{y}_1\right)$ is $\mathrm{T}=\mathrm{S}_1$</p> <p>$$\begin{aligned} & \therefore \mathrm{xx}_1-4\left(\mathrm{y}+\mathrm{y}_1\right)=\mathrm{x}_1^2-8 \mathrm{y}_1 \\ & \therefore\left(\mathrm{x}_1, \mathrm{y}_1\right)=\left(1, \frac{5}{4}\right) \\ & \Rightarrow \mathrm{x}-4\left(\mathrm{y}+\frac{5}{4}\right)=1-8 \times \frac{5}{4}=-9 \end{aligned}$$</p> <p>$\therefore x-4 y+4=0$ ...... (i)</p> <p>$(\alpha, \beta)$ lies on (i) & also on $y^2=4 x$</p> <p>$$\begin{aligned} & \therefore \alpha-4 \beta+4=0 \text{ .... (ii)} \\ & \& ~\beta^2=4 \alpha \text{ .... (iii)} \end{aligned}$$</p> <p>Solving (ii) & (iii)</p> <p>$$\begin{aligned} & \beta^2=4(4 \beta-4) \Rightarrow \beta^2-16 \beta+16=0 \\ & \therefore \beta=8 \pm 4 \sqrt{3} \text { and } \alpha=4 \beta-4=28 \pm 16 \sqrt{3} \\ & \therefore(\alpha, \quad \beta) \quad=\quad(28+16 \sqrt{3}, 8+4 \sqrt{3}) \quad \& \\ & (28-16 \sqrt{3}, 8-4 \sqrt{3}) \\ & \therefore(\alpha-28)(\beta-8)=( \pm 16 \sqrt{3})( \pm 4 \sqrt{3}) \\ & =192 \end{aligned}$$</p>