Equation of the section of a cone

Question

My purpose is to start from a plane $\pi$ of equation $z=ax+ by +c$ that intersects a cone $\kappa$ of equation $x^2+ y^2= k^2 z^2$ and find, analytically, the equation of the conic (intersection of the two surfaces), in a reference system in the plane $\pi$, so that this equation can be compared to the classical equation of a conic : $$Ax^2 + 2Bxy + Cy^2 + 2Dx+ 2Ey +F=0 \quad (1) $$ in a way that make explicit the link between the parameters $a,b,c$ of the plane $\pi$ and the matrix that gives the conic: $$ \begin{bmatrix}x&y&1\end{bmatrix} \begin{bmatrix} A&B&D\\ B&C&E\\ D&E&F \end{bmatrix} \begin{bmatrix}x\\y\\1\end{bmatrix}=0 $$

To simplify a bit, take $k=1$ so that the intersection of the plane and the cone gives: $$ \begin {cases} z=ax+by+c\\ x^2(a^2-1)+y^2(b^2-1)+2abxy+2acx+2bcy+c^2=0 \end{cases} $$ The second equation seems the searched equation, but I think that really it is the equation of the conic projected on the plane $z=0$.

If we want the equation referred to the plane $\pi$how we can proceed from this?

Answer 1

Let $\alpha$ be the given plane and $\vec n=(-a,-b,1)$ its normal vector. Let then $\theta$ and $\phi$ be the angles formed by $\vec n$ with $z$-axis and by its projection on the $xy$ plane with $x$-axis. The matrices performing rotations by an angle $\phi$ about $z$-axis and by an angle $\theta$ about $y$-axis are respectively: $$ R_1=\pmatrix{ \cos\phi & \sin\phi & 0 \\ -\sin\phi & \cos\phi & 0 \\ 0 & 0 & 1 \\ } \quad\text{and}\quad R_2=\pmatrix{ \cos\theta & 0 & -\sin\theta \\ 0 & 1 & 0 \\ \sin\theta & 0 & \cos\theta \\ }, $$ where: $$ \cos\phi=-\frac{a}{\sqrt{a^2+b^2}},\quad \sin\phi=-\frac{b}{\sqrt{a^2+b^2}},\quad \cos\theta=\frac{1}{\sqrt{1+a^2+b^2}},\quad \sin\theta=\frac{\sqrt{a^2+b^2}}{\sqrt{1+a^2+b^2}}. $$ Matrix $R=R_2\circ R_1$ represents then a rotation carrying then $\vec n$ on $z$-axis, while the intersection line between $\alpha$ and $xy$ plane is transformed to a line parallel to $y$-axis. We can compute $R$ and check that it transforms plane $-ax-by+z=c$ to plane $\displaystyle z=\frac{c}{\sqrt{1+a^2+b^2}}$, while cone $x^2+y^2=k^2z^2$ is transformed to $$ \left(a^2+b^2+1\right)y^2 + \left(1-k^2 \left(a^2+b^2\right)\right)x^2 +2 \left(k^2+1\right)\sqrt{a^2+b^2} x z + \left(a^2+b^2-k^2\right)z^2 =0. $$ Substituting here $\displaystyle z=\frac{c}{\sqrt{1+a^2+b^2}}$ one then gets the equation of the conic section: $$ \left(a^2+b^2+1\right)y^2 -\left(k^2\left(a^2+b^2\right)-1\right)x^2 +2 c \left(k^2+1\right)\sqrt{\frac{a^2+b^2}{a^2+b^2+1}} x +\frac{c^2 \left(a^2+b^2-k^2\right)}{a^2+b^2+1}=0. $$ The origin of this coordinate system on plane $\alpha$ is $O'= \displaystyle \frac{c}{\sqrt{1+a^2+b^2}}(-a,-b,1)$, that is the projection of $O$ on $\alpha$, while $y$-axis is parallel to the intersection between $\alpha$ and $xy$ plane.