Normals to the parabola $$y^2 = 4x$$ are drawn at two points P and Q on it. These normals intersect the parabola at the point R (9, -6). Then PQ equals

$$y^2 = 4x$$  , a =1

Equation of normal is $$y=mx-2am-am^3$$

=> $$y=mx-2m-m^3$$

It passes through R(9,-6)

=> -6 = 9m -2m - $$m^3$$

=> $$\left(m+1\right)\left(m-3\right)\left(m+2\right)=0$$

=> m =-1,3,-2

$$y^2 = 4x$$

$$2yy'=4\ =>\ y'=\frac{2}{y}$$

Slope of the normal = -y/2

-y/2 = -1 => y=2 => x= 1  hence one point is P(1,2)

-y/2 = 3 => y=-6 => x=9 hence secoond point is R(9,-6)

-y/2=-2 => y =4 => x=4 hence third point is Q(4,4)

PQ = $$\sqrt{\ 13}$$