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I'm not a math person or coder, so please be patient with me. If I have an ellipse using this equation: $$\frac{(x-h)^2}{a^2}+\frac{\left(y-k\right)^2}{b^2}=1$$

And I know the values $a,b,h,k,x',y'$ how do I calculate the coordinates for a point on the ellipse where the normal intersects $x',y'$?

enter image description here

I did find this solution here: How to find the point on an ellipse that is closest to the point A outside of the ellipse

But I don't understand how to use the equations, and also can't comment to ask for help because I don't have enough reputation points.

Much appreciated!

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  • $\begingroup$ Why are you trying to do a problem where everything is a letter rather than an identified number? The earlier problem had the specific $x^2 + 4 y^2$ for example. $\endgroup$ – Will Jagy May 29 '18 at 23:39
  • $\begingroup$ Because it's going in a script? I guess for simple example (h,k) can be at (0,0), b=1, a=2, and (x',y') is at (2,2), I just wanted to keep them variables so I can use it in a script later. $\endgroup$ – user27068 May 30 '18 at 0:24
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A central conic has at most $4$ concurrent normals from a given point.

Given an ellipse

$$\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1 \tag{1}$$

The equation of the normal at $(X,Y)$ is

$$\frac{(X-h)(y-Y)}{a^2}-\frac{(x-X)(Y-k)}{b^2}=0$$

If $B$, $C$, $D$ and $E$ are the feet of the concurrent normals from $A(x',y')$, then switching the roles of $(x,y)$ and $(X,Y)$ gives a rectangular hyperbola

$$\frac{(x-h)(y-y')}{a^2}-\frac{(x-x')(y-k)}{b^2}=0 \tag{2}$$

on which $A$, $B$, $C$, $D$ and $E$ are lying. Hence, the intersections of $(1)$ and $(2)$ are the required feet of the concurrent normals.

enter image description here

  • There will be $4$ concurrent normals when

    $$[a(x'-h)]^{2/3}+[b(y'-k)]^{2/3}<(a^2-b^2)^{2/3}$$

  • There will be $3$ concurrent normals when

    $$[a(x'-h)]^{2/3}+[b(y'-k)]^{2/3}=(a^2-b^2)^{2/3}$$

    that is $A$ lies on the ellipse evolute.

  • There will be $2$ concurrent normals when

    $$[a(x'-h)]^{2/3}+[b(y'-k)]^{2/3}>(a^2-b^2)^{2/3}$$

  • For $a^2>2b^2$, part of the evolute will be outside the ellipse. There'll be two to four normals concurrent at $(x',y')$. The problem can be solved graphically by finding the intersections of the two conics. The pictures are generated by Geogebra.

See my derivation for an oblique case of central conics here and also the case of hyperbola for your further interest.

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  • $\begingroup$ OP admitted the solution in the linked thread was opaque to OP, so a discussion like this (including hyperbolas, which were not even asked for) will likely not be a usable answer for OP. (And they aren't for me neither,- i do not even see the normals' equations) $\endgroup$ – bukwyrm May 30 '18 at 12:11
  • $\begingroup$ Ha, yea this is way over my head and also too complicated. I'm just looking for the solution to 'A' given here, I thought asking for the normal implied outward facing right angles because that is how it is used in 3D modeling, but I see here it means any right angle. $\endgroup$ – user27068 Jun 7 '18 at 9:13

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