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 $h\ge2$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.