Cutting a cone according to specs !!

The eccentricity e of a conic section is $e = \frac{\sin \phi}{\cos \beta}$ , where $\phi$ is the angle that the cutting plane normal makes with the axis of the cone, and $\beta$ is the semi-vertical angle. The eccentricity of an ellipse is $e = \sqrt{1 - \frac{b^2}{a^2}}$ , where $b$ is the semi-minor axis and $a$ is the semi-major axis. Therefore, $\frac{\sin \phi}{\cos \beta} = \sqrt{1 - \frac{b^2}{a^2}}$ , and in this problem, $\frac{\sin \phi}{\cos (\tan ^{-1} \frac{1}{2})} = \sqrt{1 - \frac{30^2}{50^2}}$ , which solves to $\phi \approx 45.69°$ , which is sufficient information to choose the correct answer.

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For completeness, the distance $z_0$ can be found by solving the triangles below from the side view of the conic section, where $A$ is the apex of the conic section, $B$ and $C$ are the intersection points of the ellipse and the surface of the conic section, $D$ is the intersection of the axis of the cone and the cutting plane of the ellipse, and $z_0 = AD$ :

Since $BC$ is twice the semi-major axis, $BC = 2 \cdot 50 = 100$ . Since $\angle BAC$ is twice the semi-vertical angle, $\angle BAC = 2 \cdot \tan ^{-1} \frac{1}{2} \approx 53.13°$ . Since $\angle BDC$ and $\phi$ are complementary, $\angle BDC \approx 90° - 45.69° \approx 44.31°$ . Then using straight angles and triangle sums, $\angle ADC \approx 135.69°$ , $\angle ABD \approx 109.13°$ , and $\angle ACD \approx 17.75°$ .

Then using the law of sines on $\triangle ABC$ , $\frac{\sin 17.75°}{AB} = \frac{\sin 53.13°}{100}$ , which solves to $AB \approx 38.10$ .

Finally, using the law of sines on $\triangle ABD$ , $\frac{\sin 109.13°}{z_0} = \frac{\sin 44.31°}{38.10}$ , which solves to $z_0 \approx 51.53$ .

Set the origin at the apex, and the z-axis to coincide with the cone axis. The cone has equation $4x^2+4y^2=z^2$

Intersect the cone with a plane $z=ax+b$ where $0<a<2$ and $b>0$ . Subtitute z, complete the square and rearrange terms to get $(\sqrt{4-a^2}x-\frac{ab}{\sqrt{4-a^2}})^2+4y^2=\frac{4b^2}{4-a^2}$

This is an equation of an ellipse, but it is merely the projection of the ellipse onto the x-y-plane. Let's set up coordinates $u,v$ in the plane of intersection. Because the plane has slope $a$ , when $x$ increases by $1$ , $z$ increases by $a$ and $u$ increases by $\sqrt{1+a^2}$ . We get a coordinate transformation (note that the origin is not at the centre of the ellipse, but we won't need that): $u= \sqrt{1+a^2}x, v=y$

Now $y$ and $v$ reach their extrema when $0^2+4y^2=\frac{4b^2}{4-a^2}$ , so the semi-minor axis is $30 = \frac{b}{\sqrt{4-a^2}}$ And $x$ and $u$ reach their extrema when $y=0$ : $|(4-a^2)x-ab|=2b$ These are apart by an x-distance of $\frac{4b}{4-a^2}$ which is to be multiplied by a factor $\sqrt{1+a^2}$ to get the (spatial) u-distance. The semi-major axis is half of that: $50 = \frac{2b\sqrt{1+a^2}}{4-a^2}$

We eliminate $b$ by taking the ratio of the semi axes and solve $a=\frac{8}{\sqrt{61}}$ $φ=\arctan{a} \approx 45°,6876$ $b=\frac{180}{61}\sqrt{305}\approx 51.5339$