Computing in Geographic Information Systems

Computational Geometry and Its Application to GIS 149

To solve for γ we know that the plane passes through (a, b, a

2

+ b

2

) so by

solving it gives

a

2

+ b

2

= 2a.a + 2b.b + γ

⇒ γ = −(a

2

+ b

2

) (7.14)

Thus the plane equation is

z = 2ax + 2by − (a

2

+ b

2

)

If we shift the plane upwards by some positive amount r

2

we get the plane

z = 2ax + 2by − (a

2

+ b

2

) + r

2

How does this plane intersect the paraboloid? Since the paraboloid is de-

fined by z = x

2

+ y

2

we can eliminate z giving:

x

2

+ y

2

= 2ax + 2by − (a

2

+ b

2

) + r

2

which after some simple rearrangements is equal to

(x − a)

2

+ (y − b)

2

= r

2

(7.15)

This is just a circle (Figure 7.16). Thus, we have shown that the intersection

of a plane with the paraboloid produces a space curve (which turns out to be

an ellipse), which when pro