# Finding the condition for point of intersection of three normals to a given parabola

Question: Suppose that the normals at three different points on the parabola $y^2=4x$ pass through the point (h,0). Show that h>2.

My attempt:

Equation of normal to parabola $y^2=4x$:

$$y=mx-2am-mx^3$$ $$a=1$$ the normals also pass through point $(h,0)$ $$0=mh-2m-m^3$$ $$m^3+2m-mh=0$$ Since three lines are passing through a single point and as parallel lines do not intersect, the three lines need to have three different slopes. So there must be three real and distinct values of m.

$$\Delta > 0$$ $$18abcd-4b^3d+b^2c^2-4ac^3-27a^2d^2>0$$ $$-4ac^3-27a^2d^2>0$$ $$4h^3>27(2)^2$$ $$h>3$$ I am getting a different condition for the intersection.

There is no need to delve into cubic equations, because your equation can be reduced very easily by taking out $m$: $$m(m^2+2-h)=0$$
So we have either $m=0$ or $m=\pm\sqrt{h-2}$ and therefore $h\ge2$