1. Point on the parabola $y^2=4x$

Let $P(\alpha,\beta)$ be on $y^2=4x$. Then $\beta^2=4\alpha \quad \Rightarrow \quad \alpha=\frac{\beta^2}{4}.$

2. Chord of the parabola $x^2=8y$ with given midpoint

We need the equation of the chord of the parabola $x^2=8y.$ This is of the form $x^2=4ay$ with $4a=8 \Rightarrow a=2.$

A standard result for the parabola $x^2=4ay$ is that the chord whose midpoint is $(x_1,y_1)$ has equation $T=S_1,$ where $T=xx_1-2a(y+y_1), \qquad S_1=x_1^2-4ay_1.$

Here $(x_1,y_1)=\left(1,\frac54\right)$ and $a=2$. So, $xx_1-2a(y+y_1)=x_1^2-4ay_1.$ Substitute values: $x(1)-4\left(y+\frac54\right)=1^2-8\cdot \frac54.$ $x-4y-5=1-10=-9.$ Hence, $x-4y+4=0.$ So the chord is $x=4y-4.$

3. Point $P(\alpha,\beta)$ lies on this chord also

Thus, $\alpha=4\beta-4.$ But from Step 1, $\alpha=\frac{\beta^2}{4}.$ Equating, $\frac{\beta^2}{4}=4\beta-4.$ Multiply by $4$: $\beta^2=16\beta-16.$ $\beta^2-16\beta+16=0.$

Solve: $\beta=\frac{16\pm\sqrt{256-64}}{2}=\frac{16\pm\sqrt{192}}{2}=\frac{16\pm 8\sqrt3}{2}=8\pm 4\sqrt3.$

Then $\alpha=4\beta-4=4(8\pm4\sqrt3)-4=28\pm16\sqrt3.$

So the two intersection points are: $P=(28+16\sqrt3,\;8+4\sqrt3)$ and $P=(28-16\sqrt3,\;8-4\sqrt3).$

4. Compute $(\alpha-28)(\beta-8)$

For either point,