Hyperbolic Conic Section

There is a formula for the eccentricity of the conic section resulting from a cutting plane intersecting a cone whose axis is vertical (i.e. along the z-axis).

$e = \dfrac{\sin \beta}{\sin \alpha}$

where $\beta$ is the angle that the cutting plane makes with the horizontal plane, and $\alpha$ is the angle that the curved surface makes with the horizontal plane. This is illustrated in the image below.

Note that $\alpha = \dfrac{ \pi }{2} - \theta_c$ , and that $\beta = \theta_p$ .

Hence the eccentricity of the resulting conic section is,

$e = \dfrac{ \sin( \tan^{-1} (2) ) }{\cos(\theta_c)} = \dfrac{ \dfrac{2}{\sqrt{5}} }{\dfrac{4}{5}} = \dfrac{\sqrt{5}}{2}$

Now the eccentricity $e$ is related to $a$ and $b$ (in the problem) as follows:

$e^2 = 1 + \dfrac{b^2}{a^2}$

Substituting the value of $e$ results in

$\dfrac{b}{a} = \dfrac{1}{2}$

Next we find $a$ from the intersection of the cutting plane with the cone in the $xz$ plane.

We have two lines from the cone, given by $z = \tan(\alpha) x$ and $z = - \tan(\alpha) x$ . These intersect with the line $z = \tan(\theta_p)(x - x_0)$ at $(x_1, y_1)$ and $( x_2, y_2)$ , respectively, where,

$x_1 = x_0 \tan(\theta_p) /(\tan(\theta_p) - \tan(\alpha) )$ , and

$x_2 = x_0 \tan(\theta_p) /(\tan(\theta_p) + \tan(\alpha) )$

Substituting $\tan(\theta_p) = 2$ and $\tan(\alpha) = \dfrac{4}{3}$ , we get,

$x_1 = 15$ and $x_2 = 3$ . Corresponding to this, we have,

$y_1 = x_1 \tan(\alpha) = 20$ , and $y_2 = -x_2 \tan(\alpha) = -4$ .

The distance between these two points that we just found is twice $a$ . Therefore,

$2 a = \sqrt{ (15-3)^2+(20+4)^2 }$ , i.e. $a = 6 \sqrt{5}$

Thus, $a^2 = 180,$ and $b^2 = \dfrac{1}{4} a^2 = 45$

The center is midpoint of the two points of intersection that we found.

$\text{Center} = \dfrac{1}{2} ( (15,0,20) + (3, 0,-4) ) = (9, 0,8)$

This makes the answer $180 + 45 + 9 + 8 = 242$