Parabola is $x^2=8 y$ Chord with mid point ( $x_1, y_1$ ) is $T=S_1$ $$ \begin{aligned} & \therefore \mathrm{xx}_1-4\left(\mathrm{y}^{+}+\mathrm{y}_1\right)=\mathrm{x}_1{ }^2-8 \mathrm{y}_1 \\ & \therefore\left(\mathrm{x}_1, \mathrm{y}_1\right)=\left(1, \frac{5}{4}\right) \\ & \Rightarrow \mathrm{x}-4\left(\mathrm{y}+\frac{5}{4}\right)=1-8 \times \frac{5}{4}=-9 \\ & \therefore \mathrm{x}-4 \mathrm{y}+4=0 \\ & (\alpha, \beta) \text { lies on (i) \& also on } y^2=4 x \\ & \therefore \alpha-4 \beta+4=0 \\ & \& \beta^2=4 \alpha \end{aligned} $$ Solving (ii) \& (iii) $$ \begin{aligned} & \beta^2=4(4 \beta-4) \Rightarrow \beta^2-16 \beta+16=0 \\ & \therefore \beta=8 \pm 4 \sqrt{3} \text { and } \alpha=4 \beta-4=28 \pm 16 \sqrt{3} \\ & \therefore(\alpha, \beta)=(28+16 \sqrt{3}, 8+4 \sqrt{3}) \\ & (28-16 \sqrt{3}, 8-4 \sqrt{3}) \\ & \therefore(\alpha-28)(\beta-8)=( \pm 16 \sqrt{3})( \pm 4 \sqrt{3}) \\ & =192 \end{aligned} $$