Given,
Parabola x2=8y
We know that,
Chord with mid point (x1,y1) is T=S1
xx1−8(2y+y1)=x1−28y1
⇒xx1−4(y+y1)=x1−28y1
Now, putting the point (x1,y1)=(1,45) we get,
⇒x−4(y+45)=1−8×45=−9
∴x−4y+4=0…(i)
(α,β) lies on (i) and also on y2=4x
∴α−4β+4=0…(ii)
And β2=4α…(iii)
Solving (ii) and (iii)
β2=4(4β−4)
⇒β2−16β+16=0
⇒β=8±43
And α=4β−4=28±163
∴(α,β)=(28+163,8+43) and (28−163,8−43)
Now, solving
(α−28)(β−8)=(163)(43)
⇒(α−28)(β−8)=192