Intersecting an Infinite Double Cone and a Plane:
looking at the equations

This section is optional in the Precalculus course.
There are no exercises in this section.

This section addresses a question about conics which may have arisen as you studied Identifying Conics by the Discriminant:

The purpose of this section is not to give a complete answer to this question.
Instead, the purpose is to give enough information—at a level within the grasp of an ambitious Precalculus student—to make the answer plausible.
The material in this section will be studied in greater detail in future calculus courses.

How do we know that intersections of an infinite double cone and a plane
give rise to an equation of the form ‘$\,Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$’?

We'll proceed as follows:

Derive the equation of an infinite double cone

  • three-dimensional coordinate system
    Use this $\,xyz\,$-coordinate system:
    • positive $\,x\,$ to the right
    • positive $\,z\,$ straight up
    • positive $\,y\,$ straight in
  • what we seek
    We seek an equation involving $\,x\,$, $\,y\,$ and $\,z\,$
    that is true for every point on an infinite double cone.
  • equation of line to be rotated
    Get the equation of the line that will be rotated about the $\,z\,$-axis to generate the cone:
    • In the $\,xz\,$-plane (not the $\,xy\,$-plane!) draw a line of slope $\,m\,$ through the origin (as shown at right).
      It has equation $\,z = mx\,$.
      (This is the familiar ‘$\,y = mx\,$’ with $\,z\,$ replacing $\,y\,$.)
    • A typical point (call it $\,P\,$) on this line has coordinates $\displaystyle\,(\frac zm,z)\,$.
      (Why? Solve $\,z = mx\,$ for $\,x\,$ to get $\displaystyle\, x = \frac zm\,$.)
  • rotate line
    Rotate this line about the $z$-axis to create an infinite double cone.
    When rotated, point $\,P\,$ creates a circle of radius $\displaystyle\,\frac{z}{m}\,$ in a a plane parallel to the $\,xy\,$-plane.
  • equation of circle
    Recall: the equation of a circle with radius $\,r\,$ through the origin is: $\,x^2 + y^2 = r^2\,$.
    Thus, the circle that $\,P\,$ creates has equation: $$ \begin{gather} x^2 + y^2 = \left(\frac{z}{m}\right)^2\cr x^2 + y^2 = \frac{z^2}{m^2}\cr\cr m^2x^2 + m^2y^2 = z^2\cr\cr (mx)^2 + (my)^2 = z^2 \qquad (\,\dagger\,) \end{gather} $$
  • equation of an infinite double cone
    Equation ($\,\dagger\,$) is the equation of an infinite double cone with side steepness $\,m\,$.

(the positive $\,y\,$ axis goes into the page)

Below are some WolframAlpha plots of infinite double cones.
The second and third plots look the same, but the scales are different—look carefully!



$x^2 + y^2 = z^2$
($\,m = 1\,$)

Observe that when $\,x = 4\,$ and $\,y = 0\,$,
$\,z\,$ is $\,1\cdot 4 = 4\,$.


$(3x)^2 + (3y)^2 = z^2$
($\,m = 3\,$)

Observe that when $\,x = 4\,$ and $\,y = 0\,$,
$\,z\,$ is $\,3\cdot 4 = 12\,$.


$(10x)^2 + (10y)^2 = z^2$
($\,m = 10\,$)

Observe that when $\,x = 4\,$ and $\,y = 0\,$,
$\,z\,$ is $\,10\cdot 4 = 40\,$.


Equation of a Plane in an $\,xyz\,$-coordinate system

Below are some WolframAlpha plots of planes.
Take a few moments to think about each of the normal vectors!


$x + y + z = 0$
normal vector: $\,\langle 1,1,1\rangle$

$2x + 3y - z = 5$
normal vector: $\,\langle 2,3,-1\rangle$

$x = 5$
normal vector: $\,\langle 5,0,0\rangle$

$y = 5$
normal vector: $\,\langle 0,5,0\rangle$

$z = 5$
normal vector: $\,\langle 0,0,5\rangle$
Note: To help WolframAlpha realize we want an equation in three variables, we put in very small coefficients for the missing terms.

Thus, for example, we input

x + 0.001y + 0.001z = 5

instead of just

x = 5

Intersect an Infinite Double Cone and a Plane

Next, we want to find the intersection of:

So, proceed as follows:

  1. At least one of $\,a\,$, $\,b\,$ or $\,c\,$ is nonzero.
    Assuming $\,c\ne 0\,$, solve the plane equation for $\,z\,$: $$ z = \frac{d - ax - by}{c} $$
  2. Substitute the expression from (1) into the equation of the infinite double cone: $$(mx)^2 + (my)^2 = \left( \frac{d - ax - by}{c}\right)^2 $$
  3. Simplify and re-arrange: $$ \begin{gather} (mx)^2 + (my)^2 = \left( \frac{d - ax - by}{c}\right)^2\cr\cr c^2m^2x^2 + c^2m^2y^2 = (d-ax-by)(d-ax-by)\cr\cr c^2m^2x^2 + c^2m^2y^2 = d^2 - adx - bdy - adx + a^2x^2 + abxy - bdy + abxy + b^2y^2\cr\cr c^2m^2x^2 + c^2m^2y^2 = d^2 - 2adx - 2bdy + a^2x^2 + 2abxy + b^2y^2\cr\cr x^2(c^2m^2 - a^2) + xy(-2ab) + y^2(c^2m^2 - b^2) + x(2ad) + y(2bd) - d^2 = 0 \qquad (**) \end{gather} $$
(Note: See what equations result if you assume $\,a\ne 0\,$ or $\,b\ne 0\,$.)

The Intersection of an Infinite Double Cone and a Plane
is an Equation with only $\,x^2\,$, $\,xy\,$, $\,y^2\,$, $\,x\,$, $\,y\,$, and constant terms

Look carefully at equation (**).
The only types of terms it is allowed to have are $\,x^2\,$, $\,xy\,$, $\,y^2\,$, $\,x\,$, $\,y\,$, and constant terms.
Thus, the intersection of an infinite double cone and a plane gives rise to an equation of the form $\,Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0\,$.

Note:
Equation (**) actually describes the projection onto the $\,xy\,$-plane of the intersection of the infinite double cone and the slicing plane.
It does not describe the curve in the slicing plane.
However, this derivation hopefully makes it plausible that the actual intersection has the desired form.

Here's an alternate approach, inspired by my genius husband Ray, that avoids the projection difficulty,
and shows that the actual equation is of the desired form.
This way rotates/translates the cone so that the slicing plane aligns with the $\,xy\,$ plane ($\,z = 0\,$).
An outline of the approach follows.

Alternate Approach

There are no exercises in this section.

Move on to the next lesson:
How is it that the Conic Discriminant
tells us the Type of Conic?