Describe all points that could be vertices of a cone whose intersection
with a plane is an ellipse that is in that plane and the distance between
whose center and either focus is c = 3 units.
From each solution statement, a reader can discern the problem, the answer,
and a justification for the answer.
In figures a and c the intersection of the
given cone and plane is an ellipse, because in those figures the plane
intersects all generating lines of the cone. In figure b the intersection
is a parabola, and in figure d it's a hyperbola.
The difference in distances between the coneÕs vertex
and A and B is the same as the difference in distances between a focus and
A and B. If the distance from that focus to the nearer of A and B is u,
then the focus is u + 6 from the other of A and B, so the difference of
these distances is (u + 6) Ð u = 6. In general, the vertex can be thought
of as being chosen so that the difference of distances to A and B is twice
the distance of a focus from the center.
We want a cone whose intersection with a plane is
an ellipse that is in that plane and the distance between whose center and
either focus is c = 3 units. According to the proof
of the conic section property for an ellipse, potential vertices for this
cone lie on a plane perpendicular to the plane of the ellipse and through
the foci of the ellipse, and satisfy the condition that the difference of
distances between the potential vertex and the foci of the ellipse is 2c = 2(3) = 6 units.
In the ellipse with equation
the value of a is 5 and the value of b is 3. Therefore the
value of c, the distance from the center to each focus, is
Hence by the reasoning in the preceding problem, potential vertices for
a cone whose intersection with the plane of the ellipse is the ellipse
lie on a plane perpendicular to the plane of the ellipse and through the
foci of the ellipse, and satisfy the condition that the difference of
distances between the potential vertex and the foci of the ellipse is
2c = 2(4) = 8 units.