Derive Equation for Dual Curve to Ellipse not at Origin Given Implicitly

Question

This is a second follow-up to this question. For the ellipse centered at $(h,k)$ defined implicitly as

$$ \frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}-1=0 \tag{1} $$

Please show the steps for calculating its dual curve and provide the equation for the dual curve. Do not solve by redefining the function parametrically. I know how to solve for parametric functions, but need to see the general algorithm for implicit functions.

Wikipedia provides a trivial example that does not help me.

My Attempt

At point $(p,q)$,

$$ X = \lambda \frac{2(p-h)}{a^2} \implies p = h + \left( \frac{a^2}{2\lambda} \right) X $$

$$ Y = \lambda \frac{2(q-k)}{b^2} \implies q = k + \left( \frac{b^2}{2\lambda} \right) Y $$

For incidence to occur, the inner product of the line $[X,Y]$ and point $(p,q)$ must be zero:

$$ \frac{a^2}{2\lambda}X^2 + hX + \frac{b^2}{2\lambda}Y^2 + kY = 0 $$

$$ a^2X^2 + b^2Y^2 + 2\lambda (hX + kY) = 0 \tag{2} $$

How should $\lambda$ be eliminated at this point?

Thought 1:

Can we homogenize the linear terms by $\frac{\left( hX + kY \right)}{-2\lambda}$? If so, we get

$$ a^2X^2 + b^2Y^2 - (hX+kY)^2 = 0 $$

$$ a^2X^2 + b^2Y^2 - (h^2X^2 + 2hkXY + k^2Y^2) = 0 $$

$$ (a^2-h^2)X^2 + (b^2-k^2)Y^2 - 2hkXY = 0 $$

but I'm not sure if this is the correct equation for the dual. Is there a way to check?

Thought 2:

Alternatively, can I just say that $\lambda = 1$? In this case, I get:

$$ a^2X^2+b^2Y^2+2(hX+kY)=0 $$

However, now my answer in Thought 1 is different from Thought 2.

Answer 1

The dual conics are for $ax^2+bxy+cy^2+dx+ey+f=0$ given by $$ (-\frac14e^2+cf)x^2+(\frac12de-bf)xy+(-\frac14d^2+af)y^2+(-cd+\frac12be)x+(\frac12bd-ae)y-\frac14b^2+ac=0$$ or $\begin{pmatrix}x&y&1\end{pmatrix}\begin{pmatrix}a&\frac{b}{2}&\frac{d}{2}\\\frac{b}{2}&c&\frac{e}{2}\\\frac{d}{2}&\frac{e}{2}&f\end{pmatrix}\begin{pmatrix}x\\y\\1\end{pmatrix}$ to $$\begin{pmatrix}x&y&1\end{pmatrix}\begin{pmatrix}-\frac14e^2+cf&\frac{\frac12de-bf}{2}&\frac{-cd+\frac12be}{2}\\\frac{\frac12de-bf}{2}&-\frac14d^2+af&\frac{\frac12bd-ae}{2}\\\frac{-cd+\frac12be}{2}&\frac{\frac12bd-ae}{2}&-\frac14b^2+ac\end{pmatrix}\begin{pmatrix}x\\y\\1\end{pmatrix}$$

from which you can see it's to do with the matrix of a conic and it's adjoint matrix which is the inverse when divided by the determinant.

For an explicit derivation see Perspectives on Projective Geometry by Jürgen Richter-Gebert:

A conic consists of all points p that satisfy an equation $p^TAp = 0.$ The set of all tangents to this conic can be described as $\{Ap | p^TAp = 0\}.$ We can describe this set of homogeneous coordinates for lines directly as a quadratic form by the following observation:

$p^TAp = p^TAA^{−1}Ap = p^TA^TA^{−1}Ap = (Ap)^TA^{−1}(Ap).$

The right side of the equation explains how the set of tangent lines of a conic may be directly interpreted as the zero set of a quadratic form with matrix $A^{−1}$.

Edit:

The $\lambda$-method works in Macaulay2:

R=QQ[a,b,c,d,e,f]
S=R[l]
T=S[p,q,r,X,Y,Z,MonomialOrder=>Lex]
I=ideal(X-l*(2*a*p+b*q+d*r),Y-l*(b*p+2*q*c+e*r),Z-l*(d*p+e*q+2*f*r),X*p+Y*q+Z*r)
gens gb I -- matrix {{(e^2-4*c*f)*X^2+(-2*d*e+4*b*f)*X*Y+(4*c*d-2*b*e)*X*Z+(d^2-4*a*f)*Y^2+(-2*b*d+4*a*e)*Y*Z+(b^2-4*a*c)*Z^2, (2*c*d^2-2*b*d*e+2*a*e^2+2*b^2*f-8*a*c*f)*l*r+(-2*c*d+b*e)*X+(b*d-2*a*e)*Y+(-b^2+4*a*c)*Z,(e^2-4*c*f)*l*r*X+(-d*e+2*b*f)*l*r*Y+(2*c*d-b*e)*l*r*Z-e*X*Y+2*c*X*Z+d*Y^2-b*Y*Z, (d*e-2*b*f)*l*r*X+(-d^2+4*a*f)*l*r*Y+(b*d-2*a*e)*l*r*Z-e*X^2+d*X*Y+b*X*Z-2*a*Y*Z,(2*c*d-b*e)*l*r*X+(-b*d+2*a*e)*l*r*Y+(b^2-4*a*c)*l*r*Z-2*c*X^2+2*b*X*Y-2*a*Y^2, (2*c*d-b*e)*l*q+(d*e-2*b*f)*l*r-d*Y+b*Z, (b*d-2*a*e)*l*q+(d^2-4*a*f)*l*r-d*X+2*a*Z, (b^2-4*a*c)*l*q+(b*d-2*a*e)*l*r-b*X+2*a*Y,(2*c*d^2-2*b*d*e+2*a*e^2)*q*Y^2+(2*b^2*e-8*a*c*e)*q*Y*Z+(-2*b^2*c+8*a*c^2)*q*Z^2+(d*e^2-4*c*d*f)*r*X*Y+(-b*e^2+4*b*c*f)*r*X*Z+(-d^2*e+4*a*e*f)*r*Y^2+(4*c*d^2-2*a*e^2-8*a*c*f)*r*Y*Z+(-4*b*c*d+b^2*e+4*a*c*e)*r*Z^2,(2*c*d-b*e)*q*X+(-b*d+2*a*e)*q*Y+(b^2-4*a*c)*q*Z+(d*e-2*b*f)*r*X+(-d^2+4*a*f)*r*Y+(b*d-2*a*e)*r*Z, e*l*q*X-d*l*q*Y+2*f*l*r*X-d*l*r*Z-X*Z, 2*c*l*q*X-b*l*q*Y+e*l*r*X-b*l*r*Z-X*Y, b*l*q*X-2*a*l*q*Y+d*l*r*X-2*a*l*r*Z-X^2,e*q*X*Y-2*c*q*X*Z-d*q*Y^2+b*q*Y*Z+2*f*r*X*Y-e*r*X*Z-d*r*Y*Z+b*r*Z^2, e*q*X^2-d*q*X*Y-b*q*X*Z+2*a*q*Y*Z+2*f*r*X^2-2*d*r*X*Z+2*a*r*Z^2, 2*c*q*X^2-2*b*q*X*Y+2*a*q*Y^2+e*r*X^2-d*r*X*Y-b*r*X*Z+2*a*r*Y*Z, d*l*p+e*l*q+2*f*l*r-Z,b*l*p+2*c*l*q+e*l*r-Y, 2*a*l*p+b*l*q+d*l*r-X, 2*a*p*Z-e*q*X+d*q*Y+b*q*Z-2*f*r*X+2*d*r*Z, d*p*Y-b*p*Z+e*q*Y-2*c*q*Z+2*f*r*Y-e*r*Z, 2*a*p*Y-2*c*q*X+2*b*q*Y-e*r*X+d*r*Y+b*r*Z, p*X+q*Y+r*Z}}

$$(e^2-4cf)X^2+(-2de+4bf)XY+(4cd-2be)XZ+(d^2-4af)Y^2+(-2bd+4ae)YZ+(b^2-4ac)Z^2=0$$

It even gives you via

oo // (gens I)

the combination of the generators that give this relation:

$$(X-\lambda (2ap+bq+dr))((e^2-4cf)X+(-de+2bf)Y+(2cd-be)Z)\\+(Y-\lambda (bp+2qc+er))((-de+2bf)X+(d^2-4af)Y+(-bd+2ae)Z)\\+(Z-\lambda (dp+eq+2fr))((2cd-be)X+(-bd+2ae)Y+(b^2-4ac)Z)\\ + (pX + qY + rZ)((2cd^2-2bde+2ae^2+2b^2f-8acf)\lambda)$$

Answer 2

I see now. So the general quadratic form in the projective plane is

$$ \begin{pmatrix} x & y & z \end{pmatrix} \begin{pmatrix} a & b & d \\ b & c & e \\ d & e & f \end{pmatrix} \begin{pmatrix} x \\ y \\ z\end{pmatrix} = 0 $$

which yields

$$ ax^2+cy^2+fz^2+2bxy+2dxz+2eyz=0, $$

but when we restrict ourselves to the affine plane,

$$ \begin{pmatrix} x & y & 1 \end{pmatrix} \begin{pmatrix} a & b & d \\ b & c & e \\ d & e & f \end{pmatrix} \begin{pmatrix} x \\ y \\ 1\end{pmatrix} = 0 $$

becomes

$$ ax^2+cy^2+f+2bxy+2dx+2ey=0. $$

The adjoint (signed cofactor matrix) is denoted $A^\Delta$:

$$ \begin{pmatrix} a & b & d \\ b & c & e \\ d & e & f \end{pmatrix} ^ \Delta = \begin{pmatrix} + \begin{vmatrix} c & e \\ e & f \end{vmatrix} & - \begin{vmatrix} b & e \\ d & f \end{vmatrix} & + \begin{vmatrix} b & c \\ d & e \end{vmatrix} \\ - \begin{vmatrix} b & d \\ e & f \end{vmatrix} & + \begin{vmatrix} a & d \\ d & f \end{vmatrix} & - \begin{vmatrix} a & b \\ d & e \end{vmatrix} \\ + \begin{vmatrix} b & d \\ c & e \end{vmatrix}& - \begin{vmatrix} a & d \\ b & e \end{vmatrix} & + \begin{vmatrix} a & b \\ b & c \end{vmatrix} \end{pmatrix} $$

Answer 3

First multiply through with $a^2b^2$ expand and homogenize the equation:

$$(a^2k^2+b^2h^2-a^2b^2)z^2-2a^2kyz-2b^2hxz+a^2y^2+b^2x^2=0$$

Then the system is

$$\begin{align}X-\lambda (2b^2p-2b^2hr)&=0\\ Y-\lambda (2a^2q-2a^2kr)&=0\\ Z-\lambda (2(a^2k^2+b^2h^2-a^2b^2)r-2a^2kq-2b^2hp)&=0\\ Xp+Yq+Zr&=0\end{align}$$

Solving explicitly is feasible by hand (showing the Sylvester matrix of which the resultant is the determinant):

resultant(X-l*(2*b^2*p-2*b^2*h*r),Y-l*(2*a^2*q-2*a^2*k*r),p);

$$\begin{pmatrix} 2\,a^2\,k \,l\,r-2\,a^2\,l\,q+Y\\\end{pmatrix}$$

resultant(X-l*(2*b^2*p-2*b^2*h*r),Z-l*(2*(a^2*k^2+b^2*h^2-a^2*b^2)*r-2*a^2*k*q-2*b^2*h*p),p);

$$\begin{pmatrix} -2\,b^2\, l&2\,b^2\,h\,l\,r+X\\ 2\,b^2\,h\,l&\left( -2\,a^2\,k^2-2\,b^2\,h^2+2\,a^2\,b^2\right)\,l\,r+2\,a^2\,k\,l\,q+Z \\ \end{pmatrix}$$

resultant(Y-l*(2*a^2*q-2*a^2*k*r),Z-l*(2*(a^2*k^2+b^2*h^2-a^2*b^2)*r-2*a^2*k*q-2*b^2*h*p),p);

$$\begin{pmatrix} 2\,a^2\,k \,l\,r-2\,a^2\,l\,q+Y\\\end{pmatrix}$$

resultant(X-l*(2*b^2*p-2*b^2*h*r),X*p+Y*q+Z*r,p);

$$\begin{pmatrix}-2\,b^2\, l&2\,b^2\,h\,l\,r+X\\ X&Z\,r+Y\,q\\ \end{pmatrix}$$

resultant(Y-l*(2*a^2*q-2*a^2*k*r),X*p+Y*q+Z*r,p);

$$\begin{pmatrix} 2\,a^2\,k \,l\,r-2\,a^2\,l\,q+Y\\ \end{pmatrix}$$

resultant(Z-l*(2*(a^2*k^2+b^2*h^2-a^2*b^2)*r-2*a^2*k*q-2*b^2*h*p),X*p+Y*q+Z*r,p);

$$\begin{pmatrix} 2\,b^2\,h \,l&\left(-2\,a^2\,k^2-2\,b^2\,h^2+2\,a^2\,b^2\right)\, l\,r+2\,a^2\,k\,l\,q+Z\\ X&Z\,r+Y\,q\\ \end{pmatrix}$$

Among these choose two and take resultant w.r.t $q$

resultant((2*X*a^2*k^2+2*X*b^2*h^2+2*Z*b^2*h-2*X*a^2*b^2)*l*r+(2*Y*b^2*h-2*X*a^2*k)*l*q-X*Z,2*a^2*k*l*r-2*a^2*l*q+Y,q);

$$\begin{pmatrix}\left(2\, Y\,b^2\,h-2\,X\,a^2\,k\right)\,l&\left(2\,X\,a^2\,k^2+2 \,X\,b^2\,h^2+2\,Z\,b^2\,h-2\,X\,a^2\,b^2\right)\,l\,r-X\,Z\\ -2\,a ^2\,l&2\,a^2\,k\,l\,r+Y\\ \end{pmatrix}$$

resultant((-(2*X*b^2*h+2*Z*b^2)*l*r)-2*Y*b^2*l*q-X^2,2*a^2*k*l*r-2*a^2*l*q+Y,q);

$$\begin{pmatrix} -2\,Y\,b^ 2\,l&\left(-2\,X\,b^2\,h-2\,Z\,b^2\right)\,l\,r-X^2\cr -2\,a^2\,l&2\,a^2\,k\,l\,r+Y\\ \end{pmatrix}$$

Finally the resultant w.r.t $r$

resultant(2*l*((2*Y*a^2*b^2*h*k+2*X*a^2*b^2*h^2+2*Z*a^2*b^2*h-2*X*a^4*b^2)*l*r-X*Y*a^2*k+Y^2*b^2*h-X*Z*a^2),-2*l*((b^2*(2*Y*a^2*k+2*Z*a^2)+2*X*a^2*b^2*h)*l*r+Y^2*b^2+X^2*a^2),r);

$$\begin{pmatrix} \left(4\, Y\,a^2\,b^2\,h\,k+4\,X\,a^2\,b^2\,h^2+4\,Z\,a^2\,b^2\,h-4\,X\,a^4\,b ^2\right)\,l^2&\left(-2\,X\,Y\,a^2\,k+2\,Y^2\,b^2\,h-2 \,X\,Z\,a^2\right)\,l\cr \left(-4\,Y\,a^2\,b^2\,k-4\,X\,a^2\,b^2\,h- 4\,Z\,a^2\,b^2\right)\,l^2&\left(-2\,Y^2\,b^2-2\,X^2\,a ^2\right)\,l\\ \end{pmatrix}$$

factor(%);
-8*X*a^4*b^2*(Y^2*k^2+2*X*Y*h*k+2*Y*Z*k+X^2*h^2+2*X*Z*h-Y^2*b^2-X^2*a^2+Z^2)*l^3

i.e. $Y^2k^2+2XYhk+2YZk+X^2h^2+2XZh-Y^2b^2-X^2a^2+Z^2=0.$

or in one fell swoop using the maxima CAS function eliminate

eliminate([X-l*(2*b^2*p-2*b^2*h*r),Y-l*(2*a^2*q-2*a^2*k*r),Z-l*(2*(a^2*k^2+b^2*h^2-a^2*b^2)*r-2*a^2*k*q-2*b^2*h*p),X*p+Y*q+Z*r],[p,q,r]);
[-16*Y*a^2*b^6*(Y^2*k^2+2*X*Y*h*k+2*Y*Z*k+X^2*h^2+2*X*Z*h-Y^2*b^2-X^2*a^2+Z^2)*l^4]

Setting $Z=1$ we have the dual by the equation in the standard affine in the dual projective plane

$$(h^2-a^2)X^2+2hkXY+(k^2-b^2)Y^2+2hX+2kY+1=0.$$