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I get these issues while dealing with curves, for example If i want to represent a point on a parabola $ax^2+by^2+2hxy+2gx+2fy+c=0$ $(h^2=ab)$in the form $(p,q)$, if I use ($p,\sqrt{q \text{ with respect to } p}$) to show the coordinates of a point, negative values of y cannot be represented because of root symbol. Is there a parametric form (like I know of circles) for this type of parabola so that we can show any point on a parabola?

Intelligenti pauca
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PIXEL
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2 Answers2

1

You're starting from the algebraic equation

$ a x^2 + b y^2 + 2 h xy + 2 g x + 2 f y + c =0 $

Define the vector $ r = \begin{bmatrix} x \\ y \end{bmatrix} $, then the above equation can be written as follows

$ r^T A r + B^T r + c = 0 $

where

$ A = \begin{bmatrix} a && h \\ h && b \end{bmatrix} $

$ B = \begin{bmatrix} 2 g \\ 2 f \end{bmatrix} $

The first step to find the parametric equation is to diagonalize $A$. When you do that, you will get

$ A = R D R^T $

where

$ D = \begin{bmatrix} D_{11} && 0 \\ 0 && 0 \end{bmatrix} $

Define $ u = R^T r $ , then

$ u^T D u + B^T R u + c = 0$

Let the vector $B^T R = G^T $, then the equation now reads

$ D_{11} u_1^2 + G_1 u_1 + G_2 u_2 + c = 0 $

Complete the square in $u_1$ to obtain

$ D_{11} \left( u_1 + \dfrac{G_1}{2 D_{11} } \right)^2 + G_2 u_2 + c - \dfrac{G_1^2}{4 D_{11} } = 0 $

Factor $G_2$ in the terms following it:

$ D_{11} \left( u_1 + \dfrac{G_1}{2 D_{11}} \right)^2 + G_2 \left( u_2 + \dfrac{c}{G_2} - \dfrac{G_1^2}{4 G_2 D_{11} } \right) = 0 $

Divide through by $D_{11}$, this will give you,

$ \left( u_1 + \dfrac{G_1}{2 D_{11}} \right)^2 + \dfrac{G_2}{D_{11}} \left( u_2 + \dfrac{c}{G_2} - \dfrac{G_1^2}{4 G_2 D_{11} } \right) = 0 $

Define

$u_{10} = - \dfrac{G_1}{2 D_{11}} $

$u_{20} = - \dfrac{c}{G_2} + \dfrac{G_1^2}{4 G_2 D_{11} } $

$ 4 p = - \dfrac{G_2}{D_{11}} $

Then the equation now reads

$ ( u_1 - u_{10} )^2 = 4 p ( u_2 - u_{20} ) $

Which is a parabola in $(u_1, u_2)$ plane.

To parameterize it, set

$ u_1 - u_{10} = t $

Then

$ u_2 - u_{20} = \dfrac{t^2}{4 p} $

i.e.

$(u_1, u_2) = (u_{10}, u_{20} ) + (t , \dfrac{1}{4 p} t^2 ) $

Now, recall that $ r = R u $, then the parametric equation of $r$ is

$ r = R \big( (u_{10}, u_{20} ) + (t , \dfrac{1}{4 p} t^2 ) \big) $

Define the vector $ u_0 = \begin{bmatrix} u_{10} \\ u_{20} \end{bmatrix} $, then

$ r = R u_0 + R \begin{bmatrix} t \\ \dfrac{1}{4p} t^2 \end{bmatrix} $

Let the vector $v_0 = R u_0$, and recall that

$ R = \begin{bmatrix} \cos \theta && - \sin \theta \\ \sin \theta && \cos \theta \end{bmatrix} $

Then

$ r = v_0 + t \begin{bmatrix} \cos \theta \\ \sin \theta \end{bmatrix} + \dfrac{t^2}{4p} \begin{bmatrix} - \sin \theta \\ \cos \theta \end{bmatrix} $

I've coded the above formulas into a small program and verified that it works correctly.

1

Note that $$a(ax^2+by^2+2hxy+2gx+2fy+c)=\\(ax+hy)^2+(ab-h^2)y^2+2agx+2afy+ac=\\(ax+hy)^2+2agx+2afy+ac=0,$$ since $ab=h^2.$ Assuming $a\neq 0$ and using the linked answer with $(Ax+Cy)^2+Dx+Ey+F=0,$ we get

$$(x,y)=(\frac{Ct^2-Et+CF}{AE-CD}, -\frac{At^2-Dt+AF}{AE-CD})$$ or $$(x,y)=(\frac1{2a}\frac{bt^2-2fht+abc}{fh-bg}, -\frac12\frac{t^2-2gt+ac}{af-gh})$$ For example, the general form equation for a parabola from focus $(f_x,f_y)$ and directrix $l x+m y+n=0:$ $$(x-f_x)^2+(y-f_y)^2=(lx+my+n)^2/(l^2+m^2)$$ has parametrization $$(x,y)=(\frac{lt^2-2m(f_y(l^2+m^2)+mn)t+lm^2((f_x^2+f_y^2)(l^2+m^2)-n^2)) }{2m^2(l^2+m^2)(l f_x + m f_y + n)},\frac{t^2+2(f_x(l^2+m^2)+ln)t+m^2((f_x^2+f_y^2)(l^2+m^2)-n^2)}{2m(l^2+m^2)(l f_x + m f_y + n)}).$$

Jan-Magnus Økland
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