Parametrization of intersection of a cone and a plane.

Question

I am trying to parametrize the curve given by the equations: $$x^2=2yz$$ $$\sqrt{2}x+1=y+z$$

My attempt:

I have tried to isolate $z$ in the second equation and subtitute it in the first and I get this:$$x^2-2\sqrt{2}xy+y^2+(y-1)^2=1$$ I was hoping I could complete the square, but I have not been able to do it.

I have also tried to isolate $x$ and substituting it I get: $$y^2+z^2-6zy-2y-2z+1=0$$ This has not been useful either.

Could someone give me a hint as to how to parametrize this curve?

Thanks in advance.

Answer 1

(Many thanks to @amd who has pointed errors in my previous calculations).

Take 3 points in the plane, for example

$$A=\begin{pmatrix}0\\1/2\\1/2\end{pmatrix}, \ \ B=\begin{pmatrix}\sqrt{2}\\5/2\\1/2\end{pmatrix}, \ \ C=\begin{pmatrix}\sqrt{2}\\1/2\\5/2\end{pmatrix}$$

Then write the following parametric equations :

$$M=A+u\vec{AB}+v\vec{AC} \ \ \iff \ \ \begin{cases}x&=&(u+v)\sqrt{2}\\y&=&1/2+2u\\z&=&1/2+2v\end{cases}\tag{1}$$

Then replace $x,y,z$ by these expressions in the equation of the cone giving :

$$(u+v)^2=(1/2+2u)(1/2+2v)$$

or

$$u^2 - 2uv + v^2 - u - v = 1/4\tag{2}$$

Let us consider (2) as a quadratic in variable $v$, $u$ being a parameter... Soving this quadratic, one gets an expression of $v$ as a function $v=f(u)$, or more exactly as 2 functions $v=f_1(u)$ and $v=f_2(u)$ (because the quadratic equation has two roots) which are :

$$\begin{cases}f_1(u)&=&\tfrac12(1+2u+\sqrt{2(4u+1)})\\f_2(u)&=&\tfrac12(1+2u-\sqrt{2(4u+1)})\end{cases}\tag{3}$$

Then a parametrization (see (1)) of the intersection is:

$$\begin{cases}x&=&(u + f_k(u))\sqrt{2}\\y&=&1/2+2u\\z&=&1/2+2 f_k(u)\end{cases}\tag{4}$$

(each function $f_k$ gives a parametrization of a part of the intersection curve).

Remark : Writing (1) under the form

$$(u-v)^2=u+v+1/4\tag{5}$$

Making in (5) the isometric change of coordinates

$$\begin{cases}U&=&\tfrac{\sqrt{2}}{2}(u+v)\\V&=&\tfrac{\sqrt{2}}{2}(u-v)\end{cases}$$

we transform it into :

$$2 V^2=\sqrt{2} U+\frac14$$

which the equation of a parabola.

Answer 2

Hint. Substitute for $z$ in the first equation to get $$x^2=2y(1+x\sqrt 2-y).$$ Then consider as a quadratic in $x$ and complete squares to get $$(x-y\sqrt 2)^2=2y,$$ which says that $y\ge 0.$ Thus, you can consider the first equation as a quadratic in $y,$ solve for $y$ and take only the positive solution. Then you would have $y$ entirely in terms of $x.$ Thus, if you let $x=t,$ then $y=\phi(t),$ and from the linear equation we get that $z=x\sqrt 2+1-y$ is also a function of $t.$ The range of $t$ will be given by considering the quantity under the square root sign after solving for $y$ in terms of $x.$

Answer 3

Continuing from where you left off after substituting for $z$ in the equation of the quadric surface, we can find that the quadratic part of the resulting equation is a perfect square, so we can rearrange it into $$(x-y\sqrt2)^2=2y.$$ This is the equation of a parabola.

There are many ways to parameterize it. For instance, a rational parameterization can be found by intersecting the line $y=mx$ with this parabola. Another possibility is to extend a Bézier curve beyond the interval $[0,1]$: Two points on the parabola that are easy to find are $(0,0)$ and $(0,1)$, and the tangents at these points intersect at $(-1/\sqrt2,0)$, therefore another parameterization of this parabola is $$2t(1-t)(-1/\sqrt2,0)+t^2(0,1) = (\sqrt2(t^2-t),t^2), t\in\mathbb R.$$

Yet another parameterization is obtained by computing an affine transformation that maps $Y=X^2$ onto this parabola and applying it to $(t,t^2)$: $$\begin{bmatrix}x\\y\end{bmatrix} = \begin{bmatrix}1&-\sqrt2\\0&2\end{bmatrix}^{-1} \begin{bmatrix}t\\t^2\end{bmatrix}.$$

Instead of working with the projection of the curve onto the $x$-$y$ plane, one can parameterize it directly. If you have a parameterization $s\mathbf u+t\mathbf v+\mathbf p$ of the plane, this effectively establishes an affine coordinate system for it. Taking $$M = \begin{bmatrix}\mathbf u&\mathbf v&\mathbf p\\0&0&1\end{bmatrix},$$ if the matrix of the quadric surface is $Q$, then the matrix of the intersection conic relative to this coordinate system is $M^TQM$. If we obtain a parameterization of this conic, we can map it back into $\mathbb R^3$ by multiplying by $M$.

For this problem, $$Q=\begin{bmatrix}1&0&0&0\\0&0&-1&0\\0&-1&0&0\\0&0&0&0\end{bmatrix}$$ and we can take $$M=\begin{bmatrix}1&1&\sqrt2\\\sqrt2&0&1\\0&\sqrt2&2\\0&0&1\end{bmatrix},$$ producing $$C = \begin{bmatrix}1&-1&-\sqrt2\\-1&1&0\\-\sqrt2&0&-2\end{bmatrix}.$$ The determinant of this matrix is nonzero and the determinant of the upper-left submatrix vanishes, so this is a parabola. Its equation is $$(u,v,1)C(u,v,1)^T = u^2-2uv+v^2-2\sqrt2u-2 = (u-v)^2-2(u\sqrt2+1) = 0.$$ Using the third method from above, a parameterization of the intersection curve is $$M \begin{bmatrix}1&-1&0\\2\sqrt2&0&2\\0&0&1\end{bmatrix}^{-1}\begin{bmatrix}t\\t^2\\1\end{bmatrix},$$ which works out to be $$x = \frac12(\sqrt2t^2-2t) \\ y = \frac12t^2 \\ z = \frac12(t^2-2\sqrt2t+2).$$ You can verify that this indeed satisfies the original two equations.