Chapter 8.4, Problem 69E

Solution

To graph the degenerate conics listed in the given table.

Given information:

A table of degenerate conics.

Graph:

The degenerate conics listed in the table are intersecting lines $x^2+xy=0$ , parallel lines $x^2=4$ , one line $y^2=0$ , no graph $x^2=-1$ , point $x^2+y^2=0$ and again no graph $x^2+y^2=-1$ .

The conic $x^2+xy=0$ degenerates to $x=0$ and $x+y=0$ . These two lines can be graphed as shown below.

  

A plane containing the axis of a cone intersects the cone.

The conic $x^2=4$ degenerates to $x=2$ and $x=-2$ , and can be graphed as shown below.

  

The conic $y^2=0$ is one line with equation $y=0$ and can be graphed as shown below.

  

The conic $x^2=-1$ gives two imaginary solutions $x=-i$ and $x=i$ . So, these two equations cannot be graphed on the real coordinate planes. So, its graph consists of coordinate axes only as shown below.

  

The equation $x^2+y^2=0$ is only true for the point $\left(0,0\right)$ , can be graphed as shown below.

  

This generated by a plane perpendicular to the axis of a cone that intersects the cone at the vertex.

The conic $x^2+y^2=-1$ has no graph, so the graph contains only the coordinate axes.