Vertex of $$y^2=4ax$$

The parabola is

$$y^2=4ax$$

with

$$a=2$$

Let

$$P=(2t^2,4t)$$

be the point of contact.

The equation of the normal at parameter

$$t$$

is

$$y=-tx+4t+2t^3$$

Let the normal meet the parabola again at

$$Q=(2s^2,4s)$$

Since both points lie on the same normal, substituting

$$Q$$

into the normal gives

$$4s=-2ts^2+4t+2t^3$$

$$2s=-ts^2+2t+t^3$$

$$ts^2+2s-(t^3+2t)=0$$

Since one root is

$$s=t$$

the other root is

$$s=-t-\frac{2}{t}$$

Hence

$$Q=\left(2\left(t+\frac2t\right)^2,-4\left(t+\frac2t\right)\right)$$

Now the vertex is

$$O=(0,0)$$

and

$$\angle POQ=90^\circ$$

Therefore

$$\overrightarrow{OP}\cdot\overrightarrow{OQ}=0$$

Using

$$P=(2t^2,4t)$$

and

$$Q=(2s^2,4s)$$

we get

$$4t^2s^2+16ts=0$$

$$ts(ts+4)=0$$

Since

$$t\neq0$$

and

$$s\neq0$$

$$ts=-4$$

But

$$s=-t-\frac2t$$

Therefore

$$t\left(-t-\frac2t\right)=-4$$

$$-t^2-2=-4$$

$$t^2=2$$

Since

$$P$$

is in the first quadrant,

$$t=\sqrt2$$

Hence

$$s=-2\sqrt2$$

Now

$$P=(4,4\sqrt2)$$

and

$$Q=(16,-8\sqrt2)$$

Therefore

$$PQ^2=(16-4)^2+(-8\sqrt2-4\sqrt2)^2$$

$$=12^2+(-12\sqrt2)^2$$

$$=144+288$$

$$=432$$

Hence

$$\frac{PQ^2}{12}=\frac{432}{12}$$

$$=36$$

Final Answer :

$$36$$