638 Geometry and Trigonometry
whereis a very small positive number. Therefore, the equationf (λ)=0 has three
roots,λ 1 ,λ 2 ,λ 3 , withλ 1 <a^2 <λ 2 <b^2 <λ 3 <c^2. These provide the three surfaces,
which are an ellipsoid forλ=λ 1 (Figure 84), a hyperboloid of one sheet forλ=λ 2 , and
a hyperboloid of two sheets forλ=λ 3 (Figure 85).
Figure 85To show that the surfaces are pairwise orthogonal we have to compute the angle
between the normals at an intersection point. We do this for the rootsλ 1 andλ 2 , the other
cases being similar. The normal to the ellipsoid at a point(x,y,z)is parallel to the vector
−→v
1 =(
x
a^2 −λ 1,
y
b^2 −λ 1,
z
c^2 −λ 1)
,
while the normal to the hyperboloid of one sheet is parallel to the vector
−→v
2 =(
x
a^2 −λ 2,
y
b^2 −λ 2,
z
c^2 −λ 2)
.
The dot product of these vectors is
−→v 1 ·−→v 2 = x
a^2 −λ 1·
x
a^2 −λ 2+
y
b^2 −λ 1·
y
b^2 −λ 2+
z
c^2 −λ 1·
z
c^2 −λ 2.
To prove that this is equal to 0, we use the fact that the point(x,y,z)belongs to both
quadrics, which translates into the relation
x^2
a^2 −λ 1+
y^2
b^2 −λ 1+
z^2
c^2 −λ 1=
x^2
a^2 −λ 2+
y^2
b^2 −λ 2+
z^2
c^2 −λ 2.
If we write this as
(
x^2
a^2 −λ 1
−
x^2
a^2 −λ 2)
+
(
y^2
b^2 −λ 1−
y^2
b^2 −λ 2)
+
(
z^2
c^2 −λ 1−
z^2
c^2 −λ 2