Suppose we have a circle of diameter $MN$ lying in a plane $\alpha$, and a point $V$ outside $\alpha$, which is the vertex of an oblique cone having circle $MN$ as its guiding line. I'll show below, following the derivation made by Apollonius of Perga in his treatise on conic sections, that the intersection between this cone and a plane can be described, using cartesian coordinates, by the usual equation of an ellipse.
Let's choose diameter $MN$ so that plane $VMN$ is perpendicular to $\alpha$. The cone is cut by another plane $\beta$, also perpendicular to $VMN$, intersecting the cone along a curve $APB$ (red in diagram below), where $A$ and $B$ are in particular the intersections of $\beta$ with lines $VM$ and $VN$. The more general case can be treated in a similar way, but the equation is then obtained with respect to a pair of oblique coordinate axes.
Let $P$ be any point on the curve and draw a plane through $P$ parallel to $\alpha$: it is easy to show that its intersection with the cone is a circle. Let $QR$ be the diameter of the circle parallel to $MN$. The perpendicular $PH$ from $P$ to $AB$ is also perpendicular to $MN$ and we have, by similitude in right triangle $QPR$:
$$
PH^2=QH\cdot RH.
$$
Draw now from $A$ and $B$ lines $AE$ and $BD$, parallel to $QR$. Triangles $QHA$ and $RHB$ are similar to $DBA$ and $EAB$, hence $QH=(BD/AB)AH$ and $RH=(AE/AB)BH$. Substituting that into the previous equation we get:
$$
PH^2={BD\cdot AE\over AB^2}AH\cdot BH.
$$
Finally, if $C$ is the midpoint of $AB$ we can write: $AH=AC+CH$, $BH=AC-CH$. With this substitution our equation becomes:
$$
PH^2={BD\cdot AE\over 4AC^2}(AC^2-CH^2),
$$
which can also be written as:
$$
{CH^2\over AC^2}+{4PH^2\over BD\cdot AE}=1.
$$
If we set up, in plane $\beta$, a coordinate system centred at $C$ and with $x$ axis along $AB$, then $CH$ and $PH$ are the coordinates of point $P$ and the above equation can be written as
$$
{x^2\over a^2}+{y^2\over b^2}=1,
\quad\text{where:}\quad
a=AC,\quad b=\sqrt{BD\cdot AE}/2,
$$
which is the usual equation of an ellipse with semi-axes $a$ and $b$ in cartesian coordinates.
Look at $$5x^2 - 4xy + 9y^2 +z(Ax +B y + C z + D) - 64=0$$ which has the required intersection with $z=0.$
Then we have the condition $576A^2+256AB+320B^2-41D^2-10496C=0$ for it to be a cone (the determinant of the $4\times 4$ matrix).
$$5x^2 - 4xy + 9y^2 -64\\-(41D^2-320B^2-256AB-576A^2)\,z^2/10496 +z\,(Ax +B y + D)=0$$
We find the apex, the singular point, by solving the systems of partials: $$(x,y,z)=(-(128B+576A)/(41D), -(320B+128A)/(41D), 128/D)\tag1$$ and check that it indeed lies on the surface.
It remains to make it a right circular cone with aperture $\frac{\pi}{8}.$
Shifting the apex to the origin by the obvious translation
$$(41D^2-320B^2-256AB-576A^2)z^2-10496Byz-10496Axz\\-94464y^2+41984xy-52480x^2=0\tag2$$
we see that we have a homogeneous form.
Remember $\frac{\sqrt{2+\sqrt2}}{2}=\cos(\frac{\pi}{8})=\frac{n_x x+ n_y y+n_z z}{\sqrt{x^2+y^2+z^2}}$ where $\hat{n}=(n_x,n_y,n_z)$ is a unit vector along the axis. We get $$\frac{2+\sqrt2}{4}(x^2+y^2+z^2)=(n_x x+ n_y y+n_z z)^2\tag3$$
Comparing coefficients in $(2)$ and $(3)$, where $k$ is a scaling factor, we get the following system, solving it in maxima CAS, precomputed a grobner basis in M2, all the while remembering that $$n_x^2+n_y^2+n_z^2=1$$
s:sqrt(2);
solve([n1*B+4*n3, A^2-2*A*B-B^2, 83968*k*A+s*B+209920*k*B+2*B, n2*A+4*n3, n1*A-n2*B+8*n3,2*s*A-9*s*B-1721344*k*B+4*A-18*B, 7774978036465664*k^2-19040*A*B-6272*B^2+287*D^2+430336*s+252941172736*k+1674112, 9033613312*s*k+2720*A*B+896*B^2-41*D^2-18067226624*k-146944, n2*n3-5248*k*B, 16*n1*n3+s*B+209920*k*B+2*B,4*n2^2-s-377856*k-2, n1*n2+20992*k, s*n2+83968*n1*k+209920*n2*k+2*n2, 4*n1^2-s-209920*k-2,s*n1+377856*n1*k+83968*n2*k+2*n1, s^2-2, 41*n2*D^2+3584*n3*A+3712*n3*B+20992*n1+52480*n2,41*n1*D^2+10880*n3*A+3584*n3*B+94464*n1+20992*n2,7*s*D^2+3253760*k*D^2+20992*n3^2+14*D^2+5248*s+3084648448*k+10496, 1312*B^3-85*A*D^2+198*B*D^2-94464*A+209920*B,1312*A*B^2+28*A*D^2-85*B*D^2+20992*A-94464*B, 146944*k*B^2-6724*k*D^2-340*n3^2+41*s+82, n2*B^2+4*n3*A-8*n3*B,7*s*B^2+67240*k*D^2+2952*n3^2+14*B^2-410*s-820, 2720*n3*A*B+896*n3*B^2-41*n3*D^2+5248*n2*B-94464*n3,20992*n2*k*B-s*n3-377856*n3*k-2*n3, 7904411648*n3^2*k+2205472*k*D^2+203360*n3^2-203*A*B+196*B^2-13448*s-26896,656*s*n3^2-1312*n3^2+5*A*B-2*B^2, 2720*A*B*D^2+896*B^2*D^2-41*D^4+3022848*A*B+671744*B^2-146944*D^2-110166016,6724*k*B*D^2+112*n3^2*A+116*n3^2*B-41*s*B-82*B,4299999936512*n3^4+47812050944*n3^2*B^2-988051456*n3^2*D^2-14193536*B^2*D^2+93275*D^4+30371007954944*k*D^2-740756291584*n3^2+1156743168*A*B-13918535680*B^2+334297600*D^2-185189072896*s-119750459392,1000195*k*D^4+257152*n3^2*B^2+56063*n3^2*D^2+3946396288*k*D^2+78609792*n3^2-2205472*s+2687499960320*k-4410944],[n1,n2,n3,A,B,D,k]);
And one of the solutions to this gives the positive octant apex:
n1:sqrt(13*sqrt(2)+16)/sqrt(82);
n2:((sqrt(2)-1)*sqrt(13*sqrt(2)+16))/sqrt(82);n3:sqrt(1-n1^2-n2^2);%r8:n3;
A:(sqrt(13*sqrt(2)+16)*(3*2^(3/2)-20)*sqrt(82)*%r8)/41;
B:((sqrt(2)-1)*sqrt(13*sqrt(2)+16)*(3*2^(3/2)-20)*sqrt(82)*%r8)/41;
D:(2^(5/2)*sqrt((2*(sqrt(2)-1)^2*(13*sqrt(2)+16)*(85*sqrt(2)+113)*(3*2^(3/2)-20)^2*%r8^2)/41-41*2^(9/2)-2296))/sqrt(41);
k:-(3*sqrt(2)+10)/1721344;
the above solution substituted into $(1)$ gives the apex
$$({{\sqrt{
13\,\sqrt{2}+16}\,\sqrt{35-5\,2\sqrt2}\,\left(1856-2^5\sqrt2\right)\,\sqrt{82}}\over{1681\,\sqrt{2}\,\sqrt{2344-
23\,2^5\sqrt2}}},{{\sqrt{13\,\sqrt{2}+16}\,\sqrt{35-5\,2\sqrt2}\,\left(59\,2^5\sqrt2-1920\right)\,\sqrt{82}
}\over{1681\,\sqrt{2}\,\sqrt{2344-23\,2^5\sqrt2}}},{{2^4\sqrt{82}}\over{\sqrt{2344-23\,2^5\sqrt2}}})$$
$$\approx(5.116868190800648,2.119476201505111,4.013576710399762).$$
Best Answer
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.