An oblique cone has a circular base centered at the origin of radius $5$, and an apex at $(0, 5, 20)$. A plane whose equation is $3x-4y+5z = 40$ cuts through the oblique cone, and the resulting cut is an ellipse. Find the equation of the ellipse in the form:
$ P(t) = C + w_1 \cos t + w_2 \sin t $
where $C$ is the center and $w_1$ and $w_2$ are the vectors specifying the semi-minor and semi-major axes of the ellipse.
My Attempt:
My attempt is incomplete, but the basic idea I am following is to parameterize the plane, as follows,
$ r = r_0 + V u $
and then plug this into the equation of the cone, which I don't have.
So I still have to find a way to find the algebraic (implicit) equation of the cone. And from there, it should be straight forward to write a quadratic form in $u$ which will give the two axes of the cut ellipse.
Any hints, comments or solutions are highly appreciated.
One way to get the equation of the oblique cone is to use a linear transformation that would map the oblique cone to a right cone, then use the equation for the right cone. I will use the following claims.
Given the above two claims, it follows that the equation of the oblique cone is given by $$ z = 20 - 4\sqrt{x^2 + (y - \tfrac 14 z)^2}. $$ From there, it suffices to parameterize the plane in the form $r(u,v) = r_0 + uw_1 + vw_2$ for suitable vectors $w_1,w_2$.
Outline of solution to rest of the problem:
To begin, we find a parameterization of the plane. First, we find an orthonormal basis with which to parameterize the plane. The plane is orthogonal to the vector $(3,-4,5)$. One vector orthogonal to this is $(4,3,0)$. An orthogonal direction to this within the plane can be found using the cross-product $$ (3,-4,5) \times (4,3,0) = (-15,20,25). $$ Normalizing these, we find that the vectors $v_1 = \frac 15(4,3,0)$ and $v_2 = \frac 1{5\sqrt{2}}(-3,4,5)$ form an orthogonal basis of the plane $(3,-4,5)^\perp$. One reference point on the plane of interest is given by setting $x = y = 0$ to get the point $(0,0,8)$. Thus, we have the following parameterization of the plane: $$ \mathbf r(s,t) = (0,0,8) + sv_1 + tv_2. $$ Notably, the map from $(s,t)$ to $\mathbf r(s,t)$ is an isometric embedding of $\Bbb R^2$. Thus, an ellipse within the plane corresponds to the "same" ellipse within $\Bbb R^2$.
Plugging $\mathbf r(s,t)$ into the equation for the cone, namely $$ 16x^2 + 16(y - \tfrac 14 z)^2 - (z - 20)^2 = 0 $$ yields $$ 16 s^{2} - \frac{12 \sqrt{2} s t}{5} + \frac{24 t^{2}}{5} - \frac{192 s}{5} - \frac{28 \sqrt{2} t}{5} = 80 $$ Now, denote $$ A = \pmatrix{16 & - \frac{6 \sqrt{2}}{5}\\- \frac{6 \sqrt{2}}{5} & \frac{24}{5}}, \quad b = \pmatrix{\frac{192}{5}\\\frac{28 \sqrt{2}}{5}},\quad c = \pmatrix{\frac{103}{77}\\ \frac{212 \sqrt{2}}{231}}, \quad \mathbf x = \pmatrix{s\\t}. $$ we can write the equation over $s$ and $t$ in the form $$ \mathbf x^TA \mathbf x + b^T \mathbf x - 80 = 0 $$ and, after computing $c = -\frac 12 A^{-1}b$, write this in the form $$ (\mathbf x - c)^TA(\mathbf x - c) = \frac{25600}{231} \implies\\ (\mathbf x - c)^TM(\mathbf x - c) =1, $$ where $M = \frac{231}{25600} A$. Conclude that within $\Bbb R^2$, the ellipse with the above equation has center $c$. The lengths of the major and minor axes are the reciprocals of the smaller and larger eigenvalues of $M$ (respectively), and the vectors corresponding to these axes are the associated eigenvectors. We can parameterize this ellipse with the equation $$ \pmatrix{s\\t} = \mathbf x = r + \lambda_1^{-1} \cos \tau \,u_1 + \lambda_2^{-1} \sin \tau \,u_2, $$ where $\lambda_1 > \lambda_2$ are the eigenvalues of $M$ and $u_1,u_2$ the associated eigenvectors. Plugging in the resulting formula for $s$ and $t$ into $\mathbf r(s,t)$ will yield the desired parameterization.