When the normals at two points on a parabola meet at a point on that parabola

Question

Question

Consider the parabola with equation $y^2 = 4 x$, parameterized by $(x,y) = (t^2,2t)$. If the normals at points corresponding to distinct parameter values $t_1$ and $t_2$ meet at a point on the parabola, find the relation between $t_1$ and $t_2$.

(Note from Blue: It's possible, even likely, that the question intends the intersection to be a distinct third point on the parabola.)

My Attempt:

Let the points be $(t_1^2,2t_1)$ and $(t_2^2,2t_2)$.

By taking the derivative at $t_1$ and $t_2$, I get the slopes of the normals as $-t_1$ and $-t_2$.

I then used point-slope form to get the equation of the two normals, and found the $x-$coordinate of intersection as $$x=t_1^2+t_2^2+t_1t_2+2$$

I am stuck over here.

Answer 1

So the points are $(t_1^2,2t_1)$ and $(t_2^2,2t_12)$

The slopes of the normals are $-t_1$ and $-t_2$ respectively.

The equations of the normal lines are

\begin{align} t_1x + y &= t_1^3+2t_1 \\ t_2x + y &= t_2^3+2t_2 \\ \hline (t_2-t_1)x &= (t_2^3-t_1^3) + 2(t_2-t_1) \\ x &= t_2^2 + t_2t_1 + t_1^2 + 2 \\ y &= t_1^3+2t_1 -t_1(t_2^2 + t_2t_1 + t_1^2 + 2) \\ y &= -t_1^2t_2 - t_1t_2^2 \end{align}

Since the point (x,y) needs to also be on the parabola, we need

\begin{align} y^2 &= 4x \\ (-t_1^2t_2 - t_1t_2^2)^2 &= 4(t_2^2 + t_2t_1 + t_1^2 + 2) \\ (t_1 t_2 - 2) (t_1^2 + t_2 t_1 + 2) (t_2^2 + t_1 t_2 + 2) &= 0 \end{align}

So $t_1t_2=2$ or $t_1(t_1+t_2)=-2$ or $t_2(t_1+t_2)=-2$

Note that these are pairwise incompatible (except for the last two, but that would mean both points are necessarily the same - the points cannot be vertically opposite each other); one parameter would become imaginary.

Added because of comment by 'Blue 6'.

We have $(x,y)=((t_1 + t_2)^2 + 2 - t_1t_2, -t_1t_2(t_1+t_2))$

If $t_1t_2=2$, $(x,y)=((t_1 + t_2)^2, -2(t_1+t_2))$.

If $t_1(t_1+t_2)=-2$, then $(x,y)=(t_2^2, 2t_2)$, that is, the normal line through the point $(t_1^2, 2t_1)$ on the parabola contains the point $(t_2^2, 2t_2)$ on the parabola.

If $t_2(t_1+t_2)=-2$, then $(x,y)=(t_1^2, 2t_1)$, that is, the normal line through the point $(t_2^2, 2t_2)$ on the parabola contains the point $(t_1^2, 2t_1)$ on the parabola.

So,if the normal lines to the parabola $y^2=4x$ at the distinct points $(t_1^2,2t_1)$ and $(t_2^2,2t_12)$ are to pass through a distinct third point on the parabola, then we need $t_1t_2=2$.

Answer 2

Hint: Also solve for the $y$-coordinates of the point of intersection. The equation of the parabola is $y^2=4x$. Substitute the point of intersection into the equation of the parabola.