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.
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.