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Find the number of normals to the parabola $y^2=8x$ through (2,1)

$$$$

I tried as follows:

Any normal to the parabola will be of the form $$y=mx-am^3-2am$$

Since the point (2,1) lies on the normal, it satisfies the equation of the normal. Thus $$2m^3+2m+1=0$$

The number of real roots of $m$ in the above equation would indicate the number of normals which satisfy the condition given in the question. However, I got stuck while trying to manually calculate the roots.

I would be grateful if somebody could please show me a more efficient method of solving the question. Many thanks in advance!

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  • $\begingroup$ I would say that given a parabola and a point $P$, there exists a unique normal to the parabola through $P$. If this is true, the required answer is 1 and you don't even need to know what the parabola and the point are. I know that this seems quite a strong statement, but still I would believe it's true. Can anybody give me any clue about this? $\endgroup$ – Paolo Franchi Jan 5 '16 at 19:20
  • $\begingroup$ Since a normal from a point $P$ to a curve is a point $Q$ at which light from $P$ reflects back to $P$, I would expect that your number will be one when $P$ is outside the parabola, typically three when inside, except for cases when the circle centered at $P$ and tangent to the parabola at $Q$ passes from inside to outside. Whoops, on reflection, I see that there may well be only one point when we’re inside. This requires a more careful analysis. $\endgroup$ – Lubin Jan 5 '16 at 19:20
  • $\begingroup$ @PaoloFranchi, if our point $P$ is on the axis of the parabola and sufficiently high, there are clearly three reflection points. $\endgroup$ – Lubin Jan 5 '16 at 19:21
  • $\begingroup$ Ok, I got it. Still, if the point is outside, there should be exactly one normal. Now, how can we find the locus of the points inside of the parabola through which passes exactly one normal? Do we use brute force, write down polynomial of degree 3 and find the discriminant, or is there a more elegant way? $\endgroup$ – Paolo Franchi Jan 5 '16 at 19:30
  • $\begingroup$ No, I was TOTALLY WRONG. Looking more closely yet, I see that there also are points outside with three normals. I think you really do need the discriminant of a cubic in all cases. I’m sure that this problem has been solved many times in the past 300 years, but when I get time (before midnight, anyway), I’ll write this up fully. $\endgroup$ – Lubin Jan 5 '16 at 20:14
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You have a cubic equation in $m$. You can find how many roots that equation has by looking at the discriminant

$$\Delta=18abcd-4b^3d+b^2c^2-4ac^3-27a^2d^2$$

for the equation $ax^3+bx^2+cx+d=0$. In your case, $a=2,\ b=0,\ c=2,\ d=1$. So your discriminant is

$$\Delta=0-0+0-4\cdot 2\cdot 2^3-17\cdot 2^2\cdot 1^2=-172$$

This is negative, so your equation has exactly one root. Your problem does not require actually finding the solution, so you are done (if, of course, your other work is correct).

For confirmation, you can look at the derivative of your equation:

$$\frac{dy}{dm}=6m^2+2$$

That derivative is always positive, so the function represented by $y$ is strictly increasing and there can be at most one zero. A cubic function must have at least one zero, so we know there is exactly one zero.

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This is a really interesting question, and although a solution really does involve a discriminant, the general answer is so pleasing from an esthetic standpoint, and my comments above were so wrong, that I thought I’d show you a treatment of the problem.

I’ll work with the parabola $Y=X^2/2$, chosen for no particular reason. But since all parabolas are of the same shape, this will do for all. Note that its focus is at $(0,1/2)$.

Take a general point $(x,y)$ in the plane, and ask what the points $(\xi,\xi^2/2)$ are on the parabola with the property that the line from $(x,y)$ to $(\xi,\xi^2/2)$ is orthogonal to the curve. The tangent to the curve has slope $\xi$, so the orthogonal has slope $-1/\xi$, and we ask for the $\frac{\xi^2}2-y=-\frac1\xi(\xi-x)$, thus $\xi^3-2y\xi=2x-2\xi$, so that, given $x$ and $y$, we get a cubic for $\xi$: $$ \xi^3-2(y-1)\xi-2x=0\,, $$ and this is the equation whose roots $\xi$ we want to examine qualitatively. Since the discriminant of $X^3-aX+b$ is $4a^3-27b^2$, the discriminant of our polynomial is $32(y-1)^3-108x^2$, or in other words we want to look at the curve $(Y-1)^3=\frac{27}8X^2$. This is a cusp cubic curve, note that the cusp is at $(0,1)$, twice as far from the vertex as the focus is.

The upshot? Below the curve $(Y-1)^3=\frac{27}8X^2$, there is only one normal to be drawn from $(x,y)$ to the parabola; above it, three; and on the curve, two normals to the parabola, except at the cusp, where all three coalesce, since the circle centered at $(0,1)$ of radius one has a contact with the parabola of order four.

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It's easy.. If normals are drawn from a point $(h, k)$ to a parabola $y^2 = 4ax.$

$1).$ There are $3$ normals if $h>2a$
$2).$ There is only one normal if $h<2a$

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  • $\begingroup$ This lacks detail and is not formatted well. Please show all the steps of the solution and use MathJax. $\endgroup$ – The Count Feb 7 '17 at 14:14
  • $\begingroup$ Actually the math behind this is quite lengthy........ As only this result is necessary for this problem therefore I posted only that $\endgroup$ – Palash gupta Feb 7 '17 at 14:22

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