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CONTEXT:

On the hyperbola $xy=c^2$, a possible parametrization is $P\left (cp,\frac{c}{p} \right )$. Using this, one can show with a little work that the equation of the normal is $$p^3x-py=c \left (p^4-1 \right )$$

Now given that this is a quartic polynomial in $p$, it is possible for the quartic polynomial to have 4 real roots. I managed to prove that if $y>0$, then this quartic has 4 distinct real roots if $$y<-\frac{1}{4c^2}x^3$$

Geometrically, this means that from any point satisfying the above inequalities, it is possible to construct 4 distinct normals that pass through that point.

I have attached a computer generated diagram for the case when $c=2$ at the bottom of this post.

PROBLEM:

However, I cannot actually see where 4 points would be such that their normals pass through any point in the blue region, where $y>0$. I can easily see 2 points on the curve, but definitely not 4.

QUESTION:

If my work above is correct, could somebody show me where these 4 points would be, or another diagram where 4 normals intersect at the one point?

enter image description here

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  • $\begingroup$ Taking the point $(-2,0)$ in the blue area, the $p$ equation is $-2p^3=2(p^4-1)$ which has only two real roots. Review your computation. (This occurs with other points.) $\endgroup$ – Yves Daoust Aug 9 '16 at 14:08
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Special case: $y=x$,

\begin{align*} px(p^2-1) &= c(p^4-1) \\ (p^2-1)[px-c(p^2+1)] &= 0 \\ (p-1)(p+1)(cp^2-px+c) &= 0 \\ p &= \pm 1, \frac{x \pm \sqrt{x^2-4c^2}}{2c} \end{align*}

There're four points on the hyperbola:

$$(c,c), (-c,-c), \left( \frac{x+\sqrt{x^2-4c^2}}{2}, \frac{x-\sqrt{x^2-4c^2}}{2} \right), \left( \frac{x-\sqrt{x^2-4c^2}}{2}, \frac{x+\sqrt{x^2-4c^2}}{2} \right)$$ at where their normals are concurrent at the point $(x,x)$ for $x^2>4c^2$.

In general, six normals (real and imaginary) can be drawn from a point of general position to a central quadric (surface).

Updated: general consideration

Let $(X,Y)$ be the foot of the normal of the central conic $C_{1}: \: ax^2+2bxy+c y^2=1$ that passes through the point $(x',y')$.

The tangent of the conic $C_{1}$ at $(X,Y)$ is

$$aXx+bYx+bXy+cYy=1$$

$$(aX+bY)x+(bX+cY)y=1$$

The equation of the normal is $$L: \: (aX+bY)(y-Y)-(bX+cY)(x-X)=0$$

Now put $(x,y)=(x',y')$ into $L$ gives a conic $C_{2}:$

$$(aX+bY)(Y-y')=(bX+cY)(X-x')$$

The required feet of the concurrent normals are $C_{1} \cap C_{2}$.

enter image description here

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  • $\begingroup$ Just to clarify, can you confirm that the denominators of the second set of points is $2c$? Or should it be $2$? $\endgroup$ – Trogdor Aug 9 '16 at 13:34
  • $\begingroup$ @Trogdor Corrected! $\endgroup$ – Ng Chung Tak Aug 9 '16 at 13:36
  • $\begingroup$ Thanks for your answer, I can now see it. Is this the only way that 4 real normals can be concurrent? $\endgroup$ – Trogdor Aug 9 '16 at 13:50
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    $\begingroup$ No it is not. There are points off the diagonal that produce four normals to the hyperbola. $\endgroup$ – Futurologist Aug 9 '16 at 14:05
  • $\begingroup$ @Trogdor, Answer updated! $\endgroup$ – Ng Chung Tak Aug 10 '16 at 8:46
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Here is an example with four normals.

enter image description here

From the point $(-8,-8)$, $p=-\sqrt3-2,-1,\sqrt3-2,1$.

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Without bothering to compute anything, I think that there are 4 normals to the hyperbola through some point $(x,y)$ that satisfy the condition $xy>c^2$. Just to be clear, I think this is a necessary but not sufficient conditions. To visualize these normals simply start drawing circles centered at the point $(x,y)$ and look for how many different circles will touch the hyperbola (I.e. count the number of circles centered at $(x,y)$ and tangent to the parabola). The line through the center point $(x,y)$ and a (the) point of tangency of a circle tangent to the hyperbola is a normal you are looking for. There should be three normals to the branch that is closer to the point and one normal to the branch that's farther.

The problem can also be approached by using Lagrange multipliers if you wish.

I hope I am not too far off in my argumentation.

To see that there are points off the diagonal that produce four normals to the hyperbola, take two different points on the same branch of the hyperbola and draw the normals through each point. They intersect at a common point, call it $Q$. Then there are two more normals through $Q$ because there should be one that is normal to the other branch for sure, giving you three different real solutions of your real four order polynomial equation. Hence there should be a fourth real solution and therefore a fourth normal. This is what I think.

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  • $\begingroup$ I understand your idea, but it is not at all obvious to me that there will be 4 tangential points of contact. $\endgroup$ – Trogdor Aug 9 '16 at 12:52

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