Here's the plan of attack.
Give equations that describe the cone and plane in the standard basis
Define "plane coordinates"
Find the coordinate transform matrix between the two coordinate systems
Find the equation of the conic section in plane coordinates
Rotate the coordinate system so our conic is oriented along the coordinate axes and thus find the major and minor axis lengths
Once we have the equation for the conic, there will still be a lot of work before we can get the semi major and semi minor axes, but hopefully someone else can chime in on this matter. First a preface on notation - scalars will be denoted with normal italic font, e.g $s$. Vectors will be denoted with upright (Roman) font, e.g $\mathrm{v}$. Matrices will be denoted with bold, e.g $\mathbf{M}$. With that out of the way, let's get started!
1: Cartesian equations of the plane and cone
First, the plane. Any plane in $\Bbb{R}^3$ is a level surface of a linear polynomial in $x,y,z$. What I mean by this is given $a,b,c\in\Bbb{R}$ we can represent the corresponding plane as
$$\mathcal{P}=\{(x,y,z)\in\Bbb{R}^3:ax+by+cz=d\}$$
Yes, apologies on using $a$ and $b$ - I will probably use them again later.
Now, the cone. The upward facing right angled cone is the surface corresponding to $\phi=\pi/4$, in spherical coordinates. The inverse transformation tells us that this means
$$\arccos\left(\frac{z}{\sqrt{x^2+y^2+z^2}}\right)=\frac{\pi}{4}$$
Taking the cosine of both sides we see that
$$\frac{z^2}{x^2+y^2+z^2}=\frac{1}{2}$$
So we get
$$\mathcal{C}=\{(x,y,z)\in\mathbb{R}^3:x^2+y^2-z^2=0\}$$
Now while we initially only wanted the upward part of the cone, we can see that rewriting this as
$$z^2=\pm\sqrt{x^2+y^2}$$
Means that our equation gives the lower part of the cone as well. So as long as our plane isn't vertical, i.e $c=0$, the system of equations
$$\begin{bmatrix}
ax+by+cz\\
x^{2} +y^{2} -z^{2}
\end{bmatrix} =\begin{bmatrix}
d\\
0
\end{bmatrix}$$
Which will always have at least one solution.
2: Plane coordinates
Our goal here is to create a new "basis" (this is technically an abuse of terminology) for $\mathbb{R}^3$ that we can write a vector as an affine transformation
$$x\hat{\mathrm{i}}+y\hat{\mathrm{j}}+z\hat{\mathrm{k}}=(x'\hat{\mathrm{i}}'+y'\hat{\mathrm{j}}'+z'\hat{\mathrm{k}}')+\mathrm{p}$$
Where $\mathrm{p}$ is some reference point on the plane, and the primed and unprimed basis vectors are related through via a linear transformation
$$\mathbf{M}\begin{bmatrix} \hat{\mathrm{i}}\\ \hat{\mathrm{j}}\\ \hat{\mathrm{k}} \end{bmatrix} =\begin{bmatrix} \hat{\mathrm{i}}'\\ \hat{\mathrm{j}}'\\ \hat{\mathrm{k}}' \end{bmatrix}$$
The inverse transformation between coordinate systems is
$$\begin{bmatrix} x\\ y\\ z \end{bmatrix}=\mathbf{M}^{-1}\begin{bmatrix} x'\\ y'\\ z' \end{bmatrix}+\mathrm{p}$$
We can multiply both sides by $\mathbf{M}$ to get the forward transformation:
$$\mathbf{M}\left(\begin{bmatrix} x\\ y\\ z \end{bmatrix}-\mathrm{p}\right)=\begin{bmatrix} x'\\ y'\\ z' \end{bmatrix}$$
First, we find the point $\mathrm{p}$. Though any point will do, it makes sense to use the one closest to the origin (of our original coordinate system). It turns out that this point is
$$\mathrm{p}=\frac{d}{a^2+b^2+c^2}(a,b,c)$$
This can be obtained either with simple coordinate geometry or, as I did it, with Lagrange multipliers. I can elaborate on how I did this if you wish. We let our third new basis vector be a unit vector parallel to this vector, that is
$$\hat{\mathrm{k}}'=\frac{(a,b,c)}{\sqrt{a^2+b^2+c^2}}$$
We want our new basis to be orthonormal, that is we want the other two basis vectors to satisfy the equations
$$\hat{\mathrm{i}}'\boldsymbol{\cdotp}\hat{\mathrm{k}}'=\hat{\mathrm{j}}'\boldsymbol{\cdotp}\hat{\mathrm{k}}'=\hat{\mathrm{i}}'\boldsymbol{\cdotp}\hat{\mathrm{j}}'=0$$
3: Finding the precise transformation matrix between the unit vectors
As of now, we have three equations for six (three components for both i and j) unknowns, so we need three extra conditions so we can uniquely determine our coordinate system. We'll only impose two extra conditions, namely $i'_1=j'_2$ and $j'_1=i'_2$. Letting $i'_1=1$ for now (this will be normalized later) we get the solutions
$$\hat{\mathrm{i}}'\propto \left(1,\frac{-a^2-b^2-2c^2+\sqrt{(a^2+b^2+2c^2)^2-4a^2b^2}}{2ab},\frac{-a^2+b^2+2c^2-\sqrt{(a^2+b^2+2c^2)^2-4a^2b^2}}{2ac}\right)$$
$$\hat{\mathrm{j}}'\propto \left(\frac{-a^2-b^2-2c^2+\sqrt{(a^2+b^2+2c^2)^2-4a^2b^2}}{2ab},1,\frac{a^2-b^2+2c^2+\sqrt{(a^2+b^2+2c^2)^2-4a^2b^2}}{2bc}\right)$$
For first one I'll use the abbreviations $\mu,\nu_1$ (in that order) and for the second one I'll use the abbreviations $\mu,\nu_2$ in that order. I'll also let $\kappa_1=\sqrt{1+\mu^2+\nu_1^2}$ and $\kappa_2=\sqrt{1+\mu^2+\nu_2^2}$ so we can normalize the above two expressions and write
$$\hat{\mathrm{i}}'=\frac{1}{\kappa_1}(1,\mu,\nu_1)~~;~~\hat{\mathrm{j}}'=\frac{1}{\kappa_2}(\mu,1,\nu_2)$$
The gammas can be found exactly in terms of $a,b,c$, but the expressions are rather lengthy so I won't bother posting them here. In practice you would do this numerically, not symbolically, anyway. Regardless, from the above expressions it's clear that
$$\hat{\mathrm{i}}'=\frac{1}{\kappa_1}(\hat{\mathrm{i}}+\mu\hat{\mathrm{j}}+\nu_1\hat{\mathrm{k}})~~;~~\hat{\mathrm{j}}'=\frac{1}{\kappa_2}(\mu\hat{\mathrm{i}}+\hat{\mathrm{j}}+\nu_2\hat{\mathrm{k}})$$
And of course
$$\hat{\mathrm{k}}'=\frac{a}{\sqrt{a^2+b^2+c^2}}\hat{\mathrm{i}}+\frac{b}{\sqrt{a^2+b^2+c^2}}\hat{\mathrm{j}}+\frac{c}{\sqrt{a^2+b^2+c^2}}\hat{\mathrm{k}}$$
Hence we can write the inverse transformation as
$$\begin{bmatrix} \hat{\mathrm{i}}'\\ \hat{\mathrm{j}}'\\ \hat{\mathrm{k}}' \end{bmatrix} =\underbrace{\begin{bmatrix} \frac{1}{\kappa _{1}} & \frac{\mu }{\kappa _{1}} & \frac{\nu _{1}}{\kappa _{1}}\\ \frac{\mu }{\kappa _{2}} & \frac{1}{\kappa _{2}} & \frac{\nu _{2}}{\kappa _{2}}\\ \frac{a}{\sqrt{a^{2} +b^{2} +c^{2}}} & \frac{b}{\sqrt{a^{2} +b^{2} +c^{2}}} & \frac{c}{\sqrt{a^{2} +b^{2} +c^{2}}} \end{bmatrix}}_{\mathbf{M}}\begin{bmatrix} \hat{\mathrm{i}}\\ \hat{\mathrm{j}}\\ \hat{\mathrm{k}} \end{bmatrix}$$
We can, of course, invert this matrix to get the forward transformation, but the expressions become extremely long, so I'll once again omit them. Recalling the inverse transformation, points $(x,y,z)$ sitting on the plane will have a zero $z'$ coordinate, precisely, they will be expressible as
$$\begin{bmatrix} x\\ y\\ z \end{bmatrix}=\mathbf{M}^{-1}\begin{bmatrix} x'\\ y'\\ 0 \end{bmatrix}+\mathrm{p}$$
4: Equation of the conic section
Recalling that the equation of the cone is $x^2+y^2-z^2=0$, we can plug in the RHS from the previous section into the equation for the cone to get the intersection surface between our plane and cone. We can represent our conic section as the quadric plane curve in our plane $\mathcal{P}$ using our primed plane coordinates
$$\begin{array}{l} f(x',y')=\\ +\left( (\mathbf{M}^{-1} )_{1,1} \ x'+(\mathbf{M}^{-1} )_{1,2} \ y'+\frac{da}{a^{2} +b^{2} +c^{2}}\right)^{2}\\ +\left( (\mathbf{M}^{-1} )_{2,1} \ x'+(\mathbf{M}^{-1} )_{2,2} \ y'+\frac{db}{a^{2} +b^{2} +c^{2}}\right)^{2}\\ -\left( (\mathbf{M}^{-1} )_{3,1} \ x'+(\mathbf{M}^{-1} )_{3,2} \ y'+\frac{dc}{a^{2} +b^{2} +c^{2}}\right)^{2}\\ =0 \end{array}$$
So it seems computing the third row of the inverse matrix is actually not needed. However, numeric matrix inversion is very fast in software such as Mathematica, so I won't worry about it too much.
5: Reorienting our coordinate system
Let's drop the primes to make life easier for ourselves.
We need to convert the above expression for $f$ into something of the form
$$f(x,y)=A{x}^2+Bxy+C{y}^2+D=0$$
Which will be more useful for further analysis. We can avoid a pile of algebra (useful for numerical calculations) by simply noting
$$\begin{array}{l} D=f( 0,0)\\ A=f( 1,0) -D\\ C=f( 0,1) -D\\ B=f( 1,1) -( A+C+D) \end{array}$$
Recall the standard (no tilting) equation for a conic is
$$ux^2+vy^2-1=0$$
So the goal here is to find some change of coordinates $(x,y)\to(X,Y)$ that will eliminate the mixed $XY$ terms. We want to find some polynomial function $F$ such that
$$f(x,y)=F(X,Y)\propto u X^2+vY^2-1$$
It turns out that a linear, not just affine, transformation will do - we don't need to move the origin. And we don't need to stretch the coordinate grid either, since our plane coordinates are already orthonormal. So we can simply use a rotation. So we are looking for a transformation of the form
$$\begin{bmatrix} X\\ Y \end{bmatrix} =\begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}\begin{bmatrix} x\\ y \end{bmatrix}$$
The inverse transformation is of course
$$\begin{bmatrix} x\\ y \end{bmatrix} =\begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix}\begin{bmatrix} X\\ Y \end{bmatrix}$$
Which means that
$$\begin{array}{l} F(X,Y)=\\ A( X\cos \theta +Y\sin \theta )^{2}\\ +B( X\cos \theta +Y\sin \theta )( -X\sin \theta +Y\cos \theta )\\ +C( -X\sin \theta +Y\cos \theta )^{2}\\ +D \end{array}$$
So we expand out and set the mixed term equal to zero:
$$\text{mixed coefficient}=2A\cos\theta\sin\theta+B(\cos^2\theta-\sin^2\theta)-2C\sin\theta\cos\theta=0$$
Using some double angle identities we get
$$(A-C)\sin(2\theta)+B\cos(2\theta)=0$$
Obviously if $B=0$ this whole process is unnecessary and our original coordinates are already correctly aligned. Apart from that however, we break the solution of this equation into two cases.
Case 1 - $A=C$
When $A=C$, our equation reduces to
$$B\cos(2\theta)=0$$
So our first solution is $\theta=\pi/4$. Now since $\sin(\pi/4)=\cos(\pi/4)$ we can factor them out from our expression for $F$:
$$F(X,Y)=\frac{1}{2}\left(C(X+Y)^2+B(Y+X)(Y-X)+C(X-Y)^2+2D\right)$$
Doing some algebra, we get
$$F(X,Y)=\left(C-\frac{B}{2}\right)X^2+\left(C+\frac{B}{2}\right)Y^2+D$$
So
$$F(X,Y)=-D\left(\underbrace{\frac{B-2C}{2D}}_{\equiv u}X^2+\underbrace{\frac{-B-2C}{2D}}_{\equiv v}Y^2-1\right)$$
If either $u=0$ or $v=0$ then we have a parabola. If not, then if the signs of $u$ and $v$ are different, we have a hyperbola. If both are nonzero and the same sign, then we have an ellipse or circle, and
$$"b"=\text{semi minor axis}=\beta=\sqrt{\frac{1}{\min(|u|,|v|)}}$$
$$"a"=\text{semi major axis}=\alpha=\sqrt{\frac{1}{\max(|u|,v|)}}$$
Meaning if $|u|>|v|$ our conic can be written in the form
$$\pm \frac{x^2}{\alpha^2}\pm \frac{y^2}{\beta^2}=1$$
Which perhaps is more familiar. The alpha and beta go in opposite places if $|u|<|v|$.
Case 2: $A\neq C$
In this case the equation for $\theta$ is not as simple. Recall that we want
$$(A-C)\sin(2\theta)+B\cos(2\theta)=0$$
We can restate this as
$$\tan(2\theta)=\frac{B}{C-A}$$
So
$$\theta=\frac{1}{2}\arctan\left(\frac{B}{C-A}\right)$$
Our mixed term is zero, by construction. The other coefficients are
$$X\text{ squared coefficient}=A\cos^2\theta-B\cos\theta\sin\theta+C\sin^2\theta$$
$$Y\text{ squared coefficient}=C\cos^2\theta+B\cos\theta\sin\theta+A\sin^2\theta$$
So then
$$u=\frac{A\cos^2\theta-B\cos\theta\sin\theta+C\sin^2\theta}{-D}$$
And
$$v=\frac{C\cos^2\theta+B\cos\theta\sin\theta+A\sin^2\theta}{-D}$$
Once again if either is zero, then the figure we have is a parabola. If the signs are different, we have a hyperbola. If both are nonzero with the same sign, we have an ellipse or hyperbola, and the same equations hold:
$$"b"=\text{semi minor axis}=\beta=\sqrt{\frac{1}{\min(|u|,|v|)}}$$
$$"a"=\text{semi major axis}=\alpha=\sqrt{\frac{1}{\max(|u|,|v|)}}$$
In response to comments - some examples with plots
Shown below is a graphic of the intersection between the plane $ax+by+cz=d$ with the cone $x^2+y^2-z^2=0$, using $a=0.7,b=4.1,c=1,d=-3.3$:
And now the plot of the conic of intersection $Ax^2+Bxy+Cy^2+D=0$, which we achieved through "plane coordinates" - in this case the coefficients are $A=0.934531,B=-0.291272,C=-1.57866,D=0.530046$:
Finally, we rotate our coordinate system by an amount $\theta=0.0576914$ to get the conic in the form $ux^2+vy^2-1=0$, in this case $u=-1.71572,v=2.99421$. The signs are different, hence we have a hyperbola -
This is just a rotated version of the previous plot.
A quick recap
We started with the equations for the plane and the cone, $ax+by+cz=d$ and $x^2+y^2-z^2=0$. Now we have an equation for their intersection, which I have as
$$ux^2+vy^2-1=0$$
Which perhaps you're more used to seeing as
$$\pm\frac{x^2}{"a"^2}\pm\frac{y^2}{"b"^2}=1$$
The quotation marks are because I used a and b earlier and I don't want to cause any confusion.
Slope and angle
Let's assume now that $u$ and $v$ have different signs, i.e, we have a hyperbola. How can we find the equations of the asymptotes and the angle between them? First, the equations. We can solve the equation for $y$ to get
$$y=\pm\sqrt{\frac{1-ux^2}{v}}$$
Which, since $u$ and $v$ are opposite signs, is
$$y=\pm\sqrt{\frac{1}{v}+\frac{|u|}{|v|}x^2}$$
So for large $x$, the asymptotes are the lines
$$y=\pm\sqrt{\frac{|u|}{|v|}}x$$
The line with the positive slope makes an angle of $\arctan(\sqrt{|u|/|v|})$ with the positive $x$ axis whereas the other makes an angle of $\pi-\arctan(\sqrt{|u|/|v|})$ with the positive $x$ axis meaning that the angle between them is $\phi_1=\pi-2\arctan(\sqrt{|u|/|v|})$. The other angle between them is $\phi_2=\frac{2\pi-2\phi_1}{2}$.
