Equation of a section plane in hyperbolic paraboloid Find

mwombenizhjb

mwombenizhjb

Answered question

2022-04-04

Equation of a section plane in hyperbolic paraboloid
Find the equation of a plane passing through Ox and intersecting a hyperbolic paraboloid x2py2q=2z(p>0,q>0) along a hyperbola with equal semi-axes.
My attempt: The equation of a plane passing through Ox is By+Cz=0. So z=ByC. Now substitute z=ByC into the equation of hyperbolic paraboloid x2py2q=2ByC and transform to x2p(yBqC)2q=B2qC2. What to do next? I don't understand how to find B and C if we assume x2p(yBqC)2q=B2qC2 is a hyperbola with equal semi-axes.

Answer & Explanation

arrebusnaavbz

arrebusnaavbz

Beginner2022-04-05Added 18 answers

A generic plane containing the x axis (Ox) has a unit normal vector
n=(0,cosθ,sinθ)
Two vectors (unit and mutually perpendicular) that span this plane are
v1=(1,0,0) and v2=(0,sinθ,cosθ)
Hence, on this plane, the position vector of any point is r=[v1,v2]u where u=[u1,u2]T is the coordinate vector of r with respect to v1,v2.
Hence, x=u1
y=sinθu2
z=cosθu2
Plug this into the equation of the hyperbolic paraboloid,
du12pdsin2θu22q=2cosθu2S
which is an equation of hyperbola in u1 and u2. If the semi-axes are equal then
p=dqsin2θ
Hence, the required angle θ satisfies |sinθ|=dqp
This determines four possible values for θ and correspondingly four planes.
kachnaemra

kachnaemra

Beginner2022-04-06Added 16 answers

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 p (parallel to x) and q (parallel to y).
The tilted plane forms an angle α (with tanα=BC) with xy plane, and is parallel to x axis. Hence the hyperbola on the plane has semi-axes p (parallel to x) and q|sinα| (parallel to y). You then get a right hyperbola if |sinα|=qp.

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