# Cone and ellipse equation

Given the cone $K=\{(x,y,z)\in \mathbb R^3 :x^2+y^2-z^2=0\}$ and the equation of $E=\{(x,y,z)\in \mathbb R^3 :z=my+c\}$
Construct a map that maps the intersection of $K$ and $E$ to the standard form $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ of an ellipse

I get the following for the intersection of $K$ and $E$: $x^2+y^2-(my+c)^2=0$
The standard form of the ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$.
How do I transform the ellipse given at the beginning to the standard form ?
This should work by rotating the original ellipse but I dont know how to construct the corresponding map.

Would appreciate any help.

• For the intersection to be an ellipse, you must have $-1\lt m\lt1$. For it to be non-degenerate, you must also have $c\ne0$.
– amd
Jan 12, 2017 at 17:55
• You will likely also need to translate the ellipse.
– amd
Jan 12, 2017 at 17:58

Transform

$$x^2+(1-m^2)y-mcy+c^2=0$$

$$x^2+(1-m^2)\left(y-\frac{mc}{2(1-m^2)}\right)^2+c^2-\left(\frac{mc}{2(1-m^2)}\right)^2=0$$

$$\frac{x^2}{\left(\dfrac{mc}{2(1-m^2)}\right)^2-c^2}+\frac{\left(y-\dfrac{mc}{2(1-m^2)}\right)^2}{\dfrac{\left(\dfrac{mc}{2(1-m^2)}\right)^2-c^2}{{1-m^2}}}=1$$

• you need to remove $y$ twice as I think Jan 12, 2017 at 17:34
• @SamehShenawy: thanks for the proofreading :)
– user65203
Jan 12, 2017 at 17:43
• you are welcome Jan 12, 2017 at 20:36

If we substitute for $z$ in the equation of the cone, we get an equation of the orthogonal projection of our ellipse onto the $x$-$y$ plane: $$g(x,y)=x^2+y^2-(my+c)^2=0.$$ The center of this ellipse is the projection of the center of the original ellipse. We can find the former by differentiation: $$\nabla g=\langle2x,2y-2m(my+c)\rangle=0$$ from which $x_c=0$ (as expected) and $y_c={mc\over1-m^2}$, and so $z_c=my_c+c={c\over1-m^2}$. The required transformation is then the composition of a translation to this point and a rotation about the $x$-axis through an angle of $\tan\theta=m$, so $\cos\theta=1/\sqrt{1+m^2}$ and $\sin\theta=m/\sqrt{1+m^2}$. Constructing the required transformation once you have these values is straightforward.

If the ultimate goal is to find the standard-form equation of this ellipse, however, this can be done without an explicit transformation. Observe that projection from the plane $z=my+c$ onto the $x$-$y$ plane shrinks the $y$-direction by a factor of $1/\cos\theta=\sqrt{1+m^2}$ and leaves the $x$-direction unchanged. So, if we can find the semi-axis lengths of the projected ellipse, we can recover the semi-axis lengths of the original.

Write the equation of the projected ellipse in matrix form as $$\begin{bmatrix}x&y&1\end{bmatrix}\begin{bmatrix}1&0&0\\0&1-m^2&-mc\\0&-mc&-c^2\end{bmatrix}\begin{bmatrix}x\\y\\1\end{bmatrix}=0.$$ If $S$ is the determinant of the central matrix above and $\lambda_1$ and $\lambda_2$ the eigenvalues of its upper-right $2\times2$ submatrix, then $a^2=-S/\lambda_1^2\lambda_2$ and $b^2=-S/\lambda_1\lambda_2^2$. In this case, the eigenvalues are just the diagonal elements and the determinant is $-c^2$, so we get $a=c/\sqrt{1-m^2}$ and $b=c/(1-m^2)$. (We could also have found these values by translating the equation to the center computed previously, but I find that to require more work.) Stretching $b$ by a factor of $\sqrt{1+m^2}$ to account for the slope of the plane, we have for the transformed equation of the ellipse $${(1-m^2)\over c^2}x'^2+{(1-m^2)^2\over c^2(1+m^2)}y'^2=1.$$

Note that $(1-m^2)/(1+m^2)\le1$, so the major axis of this ellipse lies along the $y'$-axis, which isn’t quite standard position. This is easily rectified, though, by swapping the coefficients of $x'^2$ and $y'^2$ in this equation. (If we’d labeled the eigenvalues so that $\lambda_1\le\lambda_2$, this issue wouldn’t have come up, but I took them in the order that they were in the matrix to preserve the $x$-axis, which keeps the mental picture a little simpler.) The rotation and translation described at the beginning of this answer have the same issue.

• Just one question. Why can we find the center of the ellipse by differentiation ? Jan 13, 2017 at 19:25
• @XPenguen It has to do with the fact that it’s the level curve of a quadratic surface. See, for example, math.stackexchange.com/q/1941210/265466 for details. This also works for hyperbolas—any conic with a center.
– amd
Jan 13, 2017 at 19:54
• We don't deal with a 3D curve directly but understood only as projection of the intersection on $x-y$ plane. . Jan 14, 2017 at 18:22
• @amd: Wouldnt we need to diagonalize the entire matrix ? I dont understand how it works the way you did it. Could you elaborate why $a^2=\frac{-S}{{\lambda_1}^2 \lambda_2}$ ? Jan 14, 2017 at 22:36
• @XPenguen It’s what you’ll end up with if you diagonalize and normalize the general matrix into the “standard” form $\operatorname{diag}(1/a^2,1/b^2,-1)$. Translations only affect the last row and column; rotations don’t change the quadratic part’s eigenvalues. Neither changes the overall determinant. After translating the general-form matrix to eliminate linear terms, you’ll have $S/\Delta=S/(\lambda_1\lambda_2)$ in the lower right corner, so you’ll need to divide by $-S/\Delta$ to bring the matrix into standard form. It’s worth going through the full calculation for yourself—once.
– amd
Jan 15, 2017 at 0:39

HINT:

Intersection of $K$ and $E$: $x^2+y^2-(my+c)^2=0$

Note that the standard form of the ellipse is $\frac{(x-0)^2}{a^2}+\frac{(y-h) ^2}{b^2}=1$.

Compare coefficients of $x^2,y^2,y$ and constants to find $h,a,b.$

• That’s not the equation of the ellipse of intersection. It’s the equation of an elliptical cylinder that is cut at an angle by the plane, so won’t have the correct eccentricity unless the plane is orthogonal to the $z$-axis.
– amd
Jan 13, 2017 at 20:13