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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.

Community
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crayon
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  • The normals to what object at points $t_1$ and $t_2$?. and what is the equation of the parabola? – Steven Alexis Gregory Jan 28 '18 at 03:55
  • @steven gregory Equation of the parabola is $y^2=4x$. The points $t_1$ and $t_2$ are on the parabola. – crayon Jan 28 '18 at 03:58
  • You should be a bit more explicit and careful in your description. You aren't merely using the equation is $y^2 = 4 x$ (with $a=1$ for simplicity). It appears that you're using the specific parameterization $(t^2, 2t)$, with $t_1$, $t_2$ being particular values of the parameter $t$. (So, $t_1$ and $t_2$ are not themselves "points", although there are points —and therefore also normals— corresponding to them.) – Blue Jan 28 '18 at 04:05
  • This question seems relevant. – amd Jan 28 '18 at 06:31

2 Answers2

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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 $t_1t_2=2$.

harry
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Steven Alexis Gregory
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    It's worth noting that the relation $t_1(t_1+t_2)=-2$ implies that the intersection of the normals is $(x,y) = (t_2^2,2t_2)$, which is one of the points determining the normals; the normal at that point, then, doesn't play much of a role in the exercise. Likewise for the relation $t_2(t_1+t_2)=-2$. Had the question indicated (or intended) that the normals meet at a distinct third point of the parabola, then the relation $t_1t_2=2$ would be the sole solution. – Blue Jan 28 '18 at 07:31
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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.

CY Aries
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