The general equation of a conic is \begin{align*}Ax^2+Bxy+Cy^2+Dx+Ey+F=0\end{align*}

### Graphing Degenerate Conics

A **degenerate conic** is a conic that does not have the usual properties of a conic. Degenerate conic equations simply cannot be written in graphing form. There are three types of degenerate conics:

**A****singular point**, which is of the form: \begin{align*}\frac{(x-h)^2}{a}+\frac{(y-k)^2}{b}=0\end{align*}(x−h)2a+(y−k)2b=0 . You can think of a singular point as a circle or an ellipse with an infinitely small radius.**A****line**, which has coefficients \begin{align*}A=B=C=0\end{align*}A=B=C=0 in the general equation of a conic. The remaining portion of the equation is \begin{align*}Dx+Ey+F=0\end{align*}Dx+Ey+F=0 , which is a line.**A****degenerate hyperbola**, which is of the form: \begin{align*}\frac{(x-h)^2}{a}-\frac{(y-k)^2}{b}=0\end{align*}(x−h)2a−(y−k)2b=0 . The result is two intersecting lines that make an “X” shape. The slopes of the intersecting lines forming the X are \begin{align*}\pm \frac{b}{a}\end{align*}±ba . This is because \begin{align*}b\end{align*}b goes with the \begin{align*}y\end{align*}y portion of the equation and is the rise, while \begin{align*}a\end{align*}a goes with the \begin{align*}x\end{align*}x portion of the equation and is the run.

### Examples

#### Example 1

Earlier, you were asked how degenerate conics are formed. When you intersect a plane with a two sided cone where the two cones touch, the intersection is a single point. When you intersect a plane with a two sided cone so that the plane touches the edge of one cone, passes through the central point and continues touching the edge of the other conic, this produces a line. When you intersect a plane with a two sided cone so that the plane passes vertically through the central point of the two cones, it produces a degenerate hyperbola.

#### Example 2

Transform the conic equation into standard form and sketch.

\begin{align*}0x^2+0xy+0y^2+2x+4y-6=0\end{align*}

This is the line \begin{align*}y=-\frac{1}{2} x+\frac{3}{2}\end{align*}

#### Example 3

Transform the conic equation into standard form and sketch.

\begin{align*}3x^2-12x+4y^2-8y+16=0\end{align*}

\begin{align*}3x^2-12x+4y^2-8y+16=0
\end{align*}

\begin{align*}3(x^2-4x)+4(y^2-2y)&=-16\\
3(x^2-4x+4)+4(y^2-2y+1)&=-16+12+4\\
3(x-2)^2+4(y-1)^2&=0\\
\frac{(x-2)^2}{4}+\frac{(y-1)^2}{3}&=0
\end{align*}

The point (2, 1) is the result of this degenerate conic.

#### Example 4

Transform the conic equation into standard form and sketch.

\begin{align*}16x^2-96x-9y^2+18y+135=0\end{align*}

\begin{align*}16x^2-96x-9y^2+18y+135=0\end{align*}

\begin{align*}16(x^2-6x)-9(y^2-2y)&=-135\\ 16(x^2-6x+9)-9(y^2-2y+1)&=-135+144-9\\ 16(x-3)^2-9(y-1)^2&=0\\ \frac{(x-3)^2}{9}-\frac{(y-1)^2}{16}&=0\end{align*}

This is a degenerate hyperbola.

#### Example 5

1. Create a conic that describes just the point (4, 7).

\begin{align*}(x-4)^2+(y-7)^2=0\end{align*}

### Review

1. What are the three degenerate conics?

Change each equation into graphing form and state what type of conic or degenerate conic it is.

2. \begin{align*}x^2-6x-9y^2-54y-72=0\end{align*}

3. \begin{align*}4x^2+16x-9y^2+18y-29=0\end{align*}

4. \begin{align*}9x^2+36x+4y^2-24y+72=0\end{align*}

5. \begin{align*}9x^2+36x+4y^2-24y+36=0\end{align*}

6. \begin{align*}0x^2+5x+0y^2-2y+1=0\end{align*}

7. \begin{align*}x^2+4x-y+8=0\end{align*}

8. \begin{align*}x^2-2x+y^2-6y+6=0\end{align*}

9. \begin{align*}x^2-2x-4y^2+24y-35=0\end{align*}

10. \begin{align*}x^2-2x+4y^2-24y+33=0\end{align*}

Sketch each conic or degenerate conic.

11. \begin{align*}\frac{(x+2)^2}{4}+\frac{(y-3)^2}{9}=0 \end{align*}

12. \begin{align*}\frac{(x-3)^2}{9}+\frac{(y+3)^2}{16}=1 \end{align*}

13. \begin{align*}\frac{(x+2)^2}{9}-\frac{(y-1)^2}{4}=1 \end{align*}

14. \begin{align*}\frac{(x-3)^2}{9}-\frac{(y+3)^2}{4}=0 \end{align*}

15. \begin{align*}3x+4y=12\end{align*}

### Review (Answers)

To see the Review answers, open this PDF file and look for section 9.6.