# Equation of a section plane in hyperbolic paraboloid Find

Equation of a section plane in hyperbolic paraboloid
Find the equation of a plane passing through Ox and intersecting a hyperbolic paraboloid $\frac{{x}^{2}}{p}-\frac{{y}^{2}}{q}=2z\left(p>0,q>0\right)$ along a hyperbola with equal semi-axes.
My attempt: The equation of a plane passing through Ox is $By+Cz=0$. So $z=\frac{-By}{C}$. Now substitute $z=\frac{-By}{C}$ into the equation of hyperbolic paraboloid $\frac{{x}^{2}}{p}-\frac{{y}^{2}}{q}=\frac{-2By}{C}$ and transform to $\frac{{x}^{2}}{p}-\frac{{\left(y-\frac{Bq}{C}\right)}^{2}}{q}=-\frac{{B}^{2}q}{{C}^{2}}$. What to do next? I don't understand how to find B and C if we assume $\frac{{x}^{2}}{p}-\frac{{\left(y-\frac{Bq}{C}\right)}^{2}}{q}=-\frac{{B}^{2}q}{{C}^{2}}$ is a hyperbola with equal semi-axes.
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arrebusnaavbz
A generic plane containing the x axis (Ox) has a unit normal vector
$n=\left(0,\mathrm{cos}\theta ,\mathrm{sin}\theta \right)$
Two vectors (unit and mutually perpendicular) that span this plane are

Hence, on this plane, the position vector of any point is $r=\left[{v}_{1},{v}_{2}\right]u$ where $u={\left[{u}_{1},{u}_{2}\right]}^{T}$ is the coordinate vector of r with respect to ${v}_{1},{v}_{2}$.
Hence, $x={u}_{1}$
$y=\mathrm{sin}\theta {u}_{2}$
$z=-\mathrm{cos}\theta {u}_{2}$
Plug this into the equation of the hyperbolic paraboloid,
$d\frac{{u}_{1}^{2}}{p}-d\frac{{\mathrm{sin}}^{2}\theta {u}_{2}^{2}}{q}=-2\mathrm{cos}\theta {u}_{2}$S
which is an equation of hyperbola in . If the semi-axes are equal then
$p=d\frac{q}{{\mathrm{sin}}^{2}\theta }$
Hence, the required angle $\theta$ satisfies $|\mathrm{sin}\theta |=\sqrt{d\frac{q}{p}}$
This determines four possible values for $\theta$ and correspondingly four planes.
###### Not exactly what you’re looking for?
kachnaemra
The last equation you got is that of a hyperbola, which is the projection of the hyperboloid section onto xy plane. The projection has semi-axes $\sqrt{p}$ (parallel to x) and $\sqrt{q}$ (parallel to y).
The tilted plane forms an angle $\alpha$ (with $\mathrm{tan}\alpha =-\frac{B}{C}$) with xy plane, and is parallel to x axis. Hence the hyperbola on the plane has semi-axes $\sqrt{p}$ (parallel to x) and $\frac{\sqrt{q}}{|\mathrm{sin}\alpha |}$ (parallel to y). You then get a right hyperbola if $|\mathrm{sin}\alpha |=\sqrt{\frac{q}{p}}$.