PlanetMath (more info)
 Math for the people, by the people. Sponsor PlanetMath
Encyclopedia | Requests | Forums | Docs | Wiki | Random | RSS  
Login
create new user
name:
pass:
forget your password?
Main Menu
Owner confidence rating: Very high Entry average rating: No information on entry rating
[parent] intersection of quadratic surface and plane (Result)

The intersection of a sphere with a plane is a circle, similarly the intersection of any surface of revolution formed by the revolution of an ellipse or a hyperbola about its axis with a plane perpendicular to the axis of revolution is a circle of latitude.

We can get as intersection curves of other quadratic surfaces and a plane also other quadratic curves (conics). If for example the ellipsoid

$\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2} \;=\; 1$ (1)

is cut with the plane $z = 0$ (i.e. the $xy$ -plane), we substitute $z = 0$ to the equation of the ellipsoid, and thus the intersection curve satisfies the equation $$\frac{x^2}{a^2}+\frac{y^2}{b^2} \;=\; 1,$$ which represents an ellipse. Actually, all plane intersections of the ellipsoid are ellipses, which may be in special cases circles.

As another exaple of quadratic surface we take the hyperbolic paraboloid

$\displaystyle \frac{x^2}{a^2}-\frac{y^2}{b^2} \;=\; 2z.$ (2)

Cutting it e.g. with the plane $y = b$ , which is parallel to the $zx$ -plane, the substitution yields the equation $$2z = \frac{x^2}{a^2}-1$$ meaning that the intersection curve in the plane $y = b$ has the projection parabola in the $zx$ -plane with such an equation, and accordingly is such a parabola.

If we cut the surface (2) with the plane $z = \frac{1}{2}$ , the result is the hyperbola having the projection $$\frac{x^2}{a^2}-\frac{y^2}{b^2} \;=\; 1$$ in the $xy$ -plane. But cutting with $z = 0$ gives $\frac{x^2}{a^2}-\frac{y^2}{b^2} \;=\; 0$ , i.e. the pair of lines $y = \pm\frac{b}{a}x$ which is a degenerate conic.

Let us then consider the general equation

$\displaystyle Ax^2+By^2+Cz^2+2A'yz+2B'zx+2C'xy+2A''x+2B''y+2C''z+D \;=\; 0$ (3)

of quadratic surface and an arbitrary plane
$\displaystyle ax+by+cz+d \;=\; 0$ (4)

where at least one of the coefficients $a$ , $b$ , $c$ is distinct from zero. Their intersection equation is obtained, supposing that e.g. $c \neq 0$ , by substituting the solved form $$z = -\frac{ax+by+d}{c}$$ of (4) to the equation (3). We then apparently have the equation of the form $$\alpha x^2+\beta y^2+2\gamma xy +2\delta x+2\varepsilon y+\zeta \;=\; 0,$$ which represents a quadratic curve or some of the degenerated cases of them.




Anyone with an account can edit this entry. Please help improve it!

"intersection of quadratic surface and plane" is owned by pahio.
(view preamble | get metadata)

View style:

See Also: quadratic surfaces, quadratic curves, conic, equation of plane, conic section


This object's parent.
Log in to rate this entry.
(view current ratings)

Cross-references: coefficients, degenerate conic, lines, projection, surface, parabola, substitution, parallel, hyperbolic paraboloid, equation, cut, ellipsoid, conics, quadratic curves, quadratic surfaces, curves, circle of latitude, axis of revolution, perpendicular, plane, axis, hyperbola, ellipse, surface of revolution, intersection, circle

This is version 3 of intersection of quadratic surface and plane, born on 2008-11-02, modified 2008-11-02.
Object id is 11231, canonical name is IntersectionOfQuadraticSurfaceAndPlane.
Accessed 1209 times total.

Classification:
AMS MSC51N20 (Geometry :: Analytic and descriptive geometry :: Euclidean analytic geometry)

Pending Errata and Addenda
None.
Discussion
Style: Expand: Order:
forum policy

No messages.

Interact
post | correct | update request | add example | add (any)