Let P(α ,β ) 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, 5 4), then (α -28)(β -8) is…

Given,

Parabola ${x}^{2}=8y$x2=8y

We know that,

Chord with mid point $({x}_{1},{y}_{1})$(x1​,y1​) is $T={S}_{1}$T=S1​

$$x{x}_{1}-8(\frac{y+{y}_{1}}{2})={x}_{1}-28{y}_{1}$$xx1​−8(2y+y1​​)=x1​−28y1​

$$\Rightarrow x{x}_{1}-4(y+{y}_{1})={x}_{1}-28{y}_{1}$$⇒xx1​−4(y+y1​)=x1​−28y1​

Now, putting the point $({x}_{1},{y}_{1})=(1,\frac{5}{4})$(x1​,y1​)=(1,45​) we get,

$$\Rightarrow x-4(y+\frac{5}{4})=1-8\times \frac{5}{4}=-9$$⇒x−4(y+45​)=1−8×45​=−9

$$\therefore x-4y+4=0\ldots (i)$$∴x−4y+4=0…(i)

$(\alpha ,\beta )$(α,β) lies on (i) and also on ${y}^{2}=4x$y2=4x

$$\therefore \alpha -4\beta +4=0\ldots (\mathrm{ii})$$∴α−4β+4=0…(ii)

And ${\beta }^{2}=4\alpha \ldots (\mathrm{iii})$β2=4α…(iii)

Solving (ii) and (iii)

$${\beta }^{2}=4(4\beta -4)$$β2=4(4β−4)

$$\Rightarrow {\beta }^{2}-16\beta +16=0$$⇒β2−16β+16=0

$$\Rightarrow \beta =8\pm 4\sqrt{3}$$⇒β=8±43​

And $\alpha =4\beta -4=28\pm 16\sqrt{3}$α=4β−4=28±163​

$\therefore (\alpha ,\beta )=(28+16\sqrt{3},8+4\sqrt{3})$∴(α,β)=(28+163​,8+43​) and $(28-16\sqrt{3},8-4\sqrt{3})$(28−163​,8−43​)

Now, solving

$$(\alpha -28)(\beta -8)=(16\sqrt{3})(4\sqrt{3})$$(α−28)(β−8)=(163​)(43​)

$$\Rightarrow (\alpha -28)(\beta -8)=192$$⇒(α−28)(β−8)=192