Normals to a Parabola

Question

A normal to the parabola at a point $A$ on the parabola whose equation is $y^{2}=2013x$ cuts the $x$-axis at $N$. $AN$ is produced to the point $B$ such that $2BN=AN$. If two more normals to the parabola $y^{2}=2013x$ pass through $B$, then prove that the angle between them is $90°$. Please provide a solution to the above question.

Answer 1

Notice first of all that instead of the given parabola we can consider the simpler equation $y^2=x$. That works because it is a mere change of scale (all parabolas are similar between them) and point $A$ is generic. Anyway one could repeat all the reasoning below for the equation $y^2=kx$, with $k$ any real number.

Take then a generic point $A=(a^2, a)$ on the parabola: the slope of the normal line at $A$ is $-2a$ and its equation is thus $y-a=-2a(x-a^2)$. The intersection between the normal and the $x$-axis can then be found to be $N=(a^2+1/2,0)$, whence: $B=(a^2+3/4,-a/2)$.

We want now to find the other normals passing through $B$. To this end, let's consider a generic point $C=(c^2,c)$ on the parabola: as before, the equation of the normal at $C$ is given by $y-c=-2c(x-c^2)$. To find for which values of $c$ this line passes through $B$ we must only substitute in that equation the coordinates of $B$ for $x$ and $y$, thus obtaining: $$ -{a\over2}-c=-2c\left(a^2+{3\over4}-c^2\right), \quad\hbox{that is:}\quad 2c^3-\left(2a^2+{1\over2}\right)c+{a\over2}=0. $$ That is a third degree equation for $c$, but we already know that $c=a$ is a solution and can factor that out. We are then left with the simple equation $$ 2c^2+2ac-{1\over2}=0. $$ This equation has two real solutions $c_1$ and $c_2$ which can readily be found: the slopes of the respective normals are then $m_1=-2c_1$ and $m_2=-2c_2$. We need only check, finally, that $m_1m_2=-1$.