Conic Section

The cutting plane passes through the point $(0, 0, z_0 )$ where $z_0 = 100$ .

We can now define a reference frame x' y' z' such that the normal vector to the cutting plane (which is $(3, 4, 5)$ ) lies in the y' z' plane. This is possible through a rotation of the x y z reference frame about the the z-axis by an angle of $\tan^{-1} (3/4)$ . In the x' y' z', the normal vector makes an angle of $\phi = \tan^{-1} (5 / \sqrt { 3^2+4^2 } ) = \pi / 4$ with the z' axis.

Next, we define a second orthogonal reference frame u v w whose origin coincides with the point (0, 0, 100), and such that the u-axis and the v-axis lie in the cutting plane, with the v-axis aligned so that it lies in the y' z' plane as well.

The unit vector along the v-axis is given by

$\hat {v }= \cos \phi \hat {j} + \sin \phi \hat {k}$

where $\hat {j}$ and $\hat {k}$ are unit vectors in the y' and z' directions.

Similarly the unit vector along the u-axis is

$\hat{u} = \hat{i}$

where $\hat{i}$ is the unit vector along the x' axis.

Next, we note that the equation of the cone is

$\vec{r} \cdot \hat{k} = (\cos \theta) r$

where $r$ is the magnitude of $\vec{r}$ .

If we consider the points of intersection between the cone and the plane, then $\vec{r}$ can be written as

$\vec{r} = z_0 \hat {k} + u \hat{u} + v \hat {v}$

Substituting the expressions for $\hat{u}$ and $\hat{v}$ we get

$\vec{r} = u \hat{i} + v \cos \phi \hat{j} + (z_0 + v \sin \phi) \hat{k}$

Using this expression for $\vec {r}$ in the defining equation of the cone above results in the following relation between u and v

$\frac{ u^2 } { a^2 } + \frac {(v -v_0)^2} {b^2} = 1$

where

$v_0 = \frac{z_0 \sin \phi \tan^2 \theta}{1 - \sin^2 \phi \sec^2 \theta}$

$a = \frac { z_0 \tan \theta \cos \phi}{\sqrt{1 - \sin^2 \phi \sec^2 \theta} }$

$b = \frac { z_0 \tan \theta \cos \phi}{1 - \sin^2 \phi \sec^2 \theta}$

Substituing $z_0 = 100, \theta = 20^{\circ} = \pi / 9 , \phi = \pi / 4$ we obtain

$a = 39.077$

$b = 59.333$

Hence

$a + b = 98.41$