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