You and your friends are playing the game Name the Conic Section. You draw a card with the equation \begin{align*}x^2  4x  8y + 12 = 0\end{align*}
Conic Equations
The equation of any conic section can be written in the form \begin{align*}Ax^2+Bxy+Cy^2+Dx+Ey+F=0\end{align*}
Standard Form of Conic Sections with Center \begin{align*}(h, k)\end{align*}(h,k)
Horizontal Axis  Vertical Axis  

Circle  \begin{align*}(xh)^2+(yk)^2=r^2\end{align*} 

Parabola  \begin{align*}(yk)^2=4p(xh)\end{align*} 
\begin{align*}(xh)^2=4p(yk)\end{align*} 
Ellipse  \begin{align*}\frac{(xh)^2}{a^2} + \frac{(yk)^2}{b^2}=1\end{align*} 
\begin{align*}\frac{(xh)^2}{b^2} + \frac{(yk)^2}{a^2}=1\end{align*} 
Hyperbola  \begin{align*}\frac{(xh)^2}{a^2}  \frac{(yk)^2}{b^2}=1\end{align*} 
\begin{align*}\frac{(yk)^2}{a^2} + \frac{(xh)^2}{b^2}=1\end{align*} 
Let's determine the type of conic section \begin{align*}x^2+y^26x+10y6=0\end{align*}
Start by rewriting the equation with the \begin{align*}x\end{align*}
\begin{align*}x^2+y^26x+10y6=0 \\
(x^26x)+(y^2+10y)=6\end{align*}
Now, complete the square for both the \begin{align*}x\end{align*}
\begin{align*}(x^26x+ {\color{blue}9}) + (y^2+10y+ {\color{red}25})&= 6+ {\color{blue}9}+ {\color{red}25} \\
(x3)^2+(y+5)^2 &= 40\end{align*}
By looking at the standard forms above, we can see that this is a circle. Another clue as to what type of conic it is, is that \begin{align*}A\end{align*}
Now, let's determine the type of conic section \begin{align*}20x^220y^280x+240y+320=0\end{align*}
Using the logic from the previous problem, we can conclude that this conic is not a circle. It is also not a parabola because it has both the \begin{align*}x^2\end{align*}
\begin{align*}20x^220y^290x+240y+320 &= 0 \\
20x^280x20y^2+240y &= 320 \\
20(x^24x)20(y^212y)&= 320\end{align*}
Now, complete the square for the \begin{align*}x\end{align*}
\begin{align*}20(x^24x)20(y^212y) &=\text{}320 \\
20(x^24x+ {\color{blue}4})20(y^212y+ {\color{red}36}) &=\text{}320+ {\color{blue}80} {\color{red}720} \\
\frac{20(x2)^2}{\text{}960}  \frac{20(y6)^2}{\text{}960} &= \frac{\text{}960}{\text{}960} \\
\text{}\frac{(x2)^2}{48} + \frac{(y6)^2}{48} &=1\\
\frac{(y6)^2}{48} \frac{(x2)^2}{48} &=1\end{align*}
We now see that this conic is a hyperbola. Going back the original equation, \begin{align*}C\end{align*}
Finally, let's write the equation of the conic below.
Just by looking at the graph, we know this is a horizontal ellipse. The standard equation for this ellipse is \begin{align*}\frac{(xh)^2}{a^2} + \frac{(yk)^2}{b^2}=1\end{align*}
Examples
Example 1
Earlier, you were asked to determine the type of conic section represented by the equation \begin{align*}x^2  4x  8y + 12 = 0\end{align*}.
Start by rewriting the equation with the \begin{align*}x\end{align*} terms and \begin{align*}y\end{align*} terms next to each other.
\begin{align*}x^24x = 8y 12\end{align*}
Now, complete the square for the \begin{align*}x\end{align*} terms. To complete the square, we need to add 4 to both sides of the equation.
\begin{align*}(x^24x+ {\color{blue}4}) = 8y 12 + {\color{blue}4}\\ (x2)^2 = 8y8\end{align*}
Finally, factor out the LCD from the right side of the equation.
\begin{align*}(x2)^2 = 8(y1)\end{align*}
By looking at the standard forms above, we can see that this is a parabola.
For Examples 2 & 3, determine the type of conic section and rewrite each equation in standard form.
Example 2
\begin{align*}9x^2+16y^2+18x135=0\end{align*}
Complete the square. Ellipse.
\begin{align*}9x^2+16y^2+18x135 &=0 \\ 9x^2+18x+16y^2 &=135 \\ 9(x^2+2x+ {\color{red}1})+16y^2 &=135+ {\color{red}9} \\ 9(x+1)^2+16y^2 &=144 \\ \frac{(x+1)^2}{16} + \frac{y^2}{9} &=1\end{align*}
Example 3
\begin{align*}y^23x8y+10=0\end{align*}
Complete the square. Parabola.
\begin{align*}y^23x8y+10 &=0 \\ y^28y3x &=10 \\ y^28y+ {\color{red}16} &=3x10+{\color{red}16} \\ (y4)^2 &=3x+6 \\ (y4)^2 &=3(x+2)\end{align*}
Example 4
Write the equation of the conic below.
This is a circle because the distance around the center is the same. The center is \begin{align*}(0, 4)\end{align*} and the radius is 5. The equation is \begin{align*}x^2+(y+4)^2=25\end{align*}.
Review
 In the general conic equation, why does B have to equal zero in order to create a conic?
Find the equation of each conic section below.
Rewrite each equation in standard form, classify the conic, and find the center. If the conic is a parabola, find the vertex.
 \begin{align*}3x^2+3y^26x+9y14=0\end{align*}
 \begin{align*}6x^2+12xy+15=0\end{align*}
 \begin{align*}x^2+2y^2+4x+2y27=0\end{align*}
 \begin{align*}x^2y^2+3x2y43=0\end{align*}
 \begin{align*}y^28x6y+49=0\end{align*}
 \begin{align*}64x^2+225y^2256x3150y3631=0\end{align*}
Answers for Review Problems
To see the Review answers, open this PDF file and look for section 10.10.