A right circular cone

Label the cross-section on the $xz$ -plane as follows, and let $AC$ bisect $\angle BAD$ :

By the distance formula, $AB = \sqrt{(10 - -5)^2+ (20 - 0)^2} = 25$ , $AD = \sqrt{(10 - 5)^2+ (20 - 0)^2} = 5\sqrt{17}$ , and $BD = \sqrt{(5 - -5)^2+ (0 - 0)^2} = 10$ .

Since $BD = 10$ , let $CD = x$ and $BC = 10 - x$ . Then by the angle bisector theorem, $\frac{CD}{BC} = \frac{AD}{AB}$ , or $\frac{x}{10-x} = \frac{5\sqrt{17}}{25}$ , which solves to $x = \frac{1}{4}(25\sqrt{17} - 85)$ .

By the Pythagorean Theorem on $\triangle ACE$ , $AC = \sqrt{(5 + \frac{1}{4}(25\sqrt{17} - 85))^2 + 20^2} = \frac{5}{4}\sqrt{10(85 - 13\sqrt{17})}$ .

By the law of cosines on $\triangle ACD$ , $\cos \alpha = \frac{CD^2 + AC^2 - AD^2}{2 \cdot CD \cdot AC} = \sqrt{\frac{1}{170}(85 - 13\sqrt{17})}$ and $\cos \beta = \frac{AC^2 + AD^2 - CD^2}{2 \cdot AC \cdot AD} = \sqrt{\frac{1}{2} + \frac{19}{10\sqrt{17}}}$ .

The conic section has an eccentricity of $\frac{\cos \alpha}{\cos \beta} = \frac{\sqrt{\frac{1}{170}(85 - 13\sqrt{17})}}{\sqrt{\frac{1}{2} + \frac{19}{10\sqrt{17}}}} = \frac{1}{2}(5 - \sqrt{17})$ , which is also equal to $\frac{\sqrt{a^2 - b^2}}{a}$ . In this case, $a = 5$ , and $b = a$ , so $\frac{1}{2}(5 - \sqrt{17}) = \frac{\sqrt{5^2 - a^2}}{5}$ , which solves to $a = 5\sqrt{\frac{1}{2}(5\sqrt{17} - 19)} \approx \boxed{4.494}$ .

The vector equation of a right cone with semi-vertical angle $\theta$ , apex at position vector $\bold{a}$ and axis parallel to a unit vector $\bold{d}$ is $(\bold{r}-\bold{a})\cdot \bold{d}=|\bold{r}-\bold{a}|\cos \theta$

(thanks to this article for the formula, which I've adapted slightly)

Here, we have $\bold{r}=(x,y,z)^T$ , $\bold{a} =(10,0,20)^T$ , and we'll put $\bold{d}=(u,v,w)^T$ . Substituting in, the equation of the cone is $u(x-10)+vy+w(z-20)=\sqrt{(x-10)^2+y^2+(z-20)^2} \cos \theta$

To find the intersection with the $xy$ -plane, put in $z=0$ : $u(x-10)+vy-20w=\sqrt{(x-10)^2+y^2+400} \cos \theta$

Squaring both sides and rearranging, the equation of the section is $(u(x-10)+vy-20w)^2 - \left((x-10)^2+y^2+400 \right) \cos^2 \theta = 0$

This has to be equivalent to the given form, though it may be a scalar multiple of it; letting $K$ be that scalar, this means $(u(x-10)+vy-20w)^2 - \left((x-10)^2+y^2+400 \right) \cos^2 \theta = K \left( \frac{x^2}{25}+\frac{y^2}{a^2}-1 \right)$

Comparing coefficients of $xy$ terms. we find $2uv=0$ ; if $u=0$ , we can't eliminate the $x^1$ term on the left-hand side, so we must have $v=0$ .

Comparing the remaining coefficients, we get the following system of equations: $\begin{aligned} u^2-\cos^2 \theta &= \frac{K}{25} \\ -2u(10u+20w)+20\cos^2 \theta &=0 \\ 100u^2+400uw+400w^2-500\cos^2 \theta &= -K \\ -\cos^2 \theta &= \frac{K}{a^2} \end{aligned}$

and $u^2+w^2=1$ , since $\bold{d}$ is a unit vector. These can be solved by progressively eliminating variables; in the end we find a single positive root for $a$ , which is $a=5 \sqrt{\frac12 (-19 + 5 \sqrt{17})} \approx \boxed{4.49378}$ .

First we note that the plane of the cut is perpendicular to the plane containing the major axis of the ellipse and the apex.

Since $A$ lies in the $xz$ plane, then the major axis of the ellipse is the one along the $x$ -axis. This way, the plane containing the $x$ axis and the apex $A$ , which is the $x z$ plane will be perpendicular to the $xy$ plane (which is the plane of the cut), while the plane containing the $y$ axis and point $A$ is not perpendicular to the $xy$ plane.

Thus the semi-major axis length is $5$ .

The cone axis will be the angle bisector in the $xz$ plane of $\angle BAC$ , where $B = (-5, 0, 0)$ and $C = (5, 0, 0)$ and $A = (10, 0, 20)$

$\angle BAC = \cos^{-1} ( \dfrac{\vec{AB} \cdot \vec{AC} } { |\vec{AB}||\vec{AC}| } )$

Hence, the semi-vertical angle $\theta = \frac{1}{2} \angle BAC$ . Next we want to find the point of intersection of this axis with the $xy$ plane. Let this point be $I$ , then we know from the angle bisector theorem that

$\dfrac{\overline{IB}}{\overline{IC}} = \dfrac{ \overline{AB}}{\overline{AC}}$

Thus, if $I = (x, 0)$ , then,

$\dfrac{(x + 5)}{(5 - x)}= \dfrac{\overline{AB}}{\overline{AC}}$

And this gives us the value of $x$ , and hence the point $I$ .

So now we have determined the axis as the vector $\vec{AI}$ , so $z_0 = | \vec{AI} |$ , and $\phi = \cos^{-1} ( \dfrac{- \hat{k} \cdot \vec{AI} }{|\vec{ A I} | } )$

Finally, we plug the obtained values in the semi-minor axis length formula, and this gives $a \approx \boxed{4.494}$