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*}. What type of conic section is represented by this equation?
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*}, which is the general second-degree equation in terms of \begin{align*}x\end{align*} and \begin{align*}y\end{align*}. For all the conic sections you have seen so far, \begin{align*}B=0\end{align*} because all axes were either horizontal or vertical. When a conic is in this form, you have to complete the square to get it into standard form.
Standard Form of Conic Sections with Center \begin{align*}(h, k)\end{align*}
Horizontal Axis | Vertical Axis | |
---|---|---|
Circle | \begin{align*}(x-h)^2+(y-k)^2=r^2\end{align*} | |
Parabola | \begin{align*}(y-k)^2=4p(x-h)\end{align*} | \begin{align*}(x-h)^2=4p(y-k)\end{align*} |
Ellipse | \begin{align*}\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2}=1\end{align*} | \begin{align*}\frac{(x-h)^2}{b^2} + \frac{(y-k)^2}{a^2}=1\end{align*} |
Hyperbola | \begin{align*}\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2}=1\end{align*} | \begin{align*}\frac{(y-k)^2}{a^2} + \frac{(x-h)^2}{b^2}=1\end{align*} |
Let's determine the type of conic section \begin{align*}x^2+y^2-6x+10y-6=0\end{align*} and rewrite the equation in standard form.
Start by rewriting the equation with the \begin{align*}x\end{align*} terms and \begin{align*}y\end{align*} terms next to each other and moving the constant to the other side of the equation.
\begin{align*}x^2+y^2-6x+10y-6=0 \\ (x^2-6x)+(y^2+10y)=6\end{align*}
Now, complete the square for both the \begin{align*}x\end{align*} and \begin{align*}y\end{align*} terms. To complete the square, you need to add \begin{align*}\left(\frac{b}{2}\right)^2\end{align*} to both sides of the equation.
\begin{align*}(x^2-6x+ {\color{blue}9}) + (y^2+10y+ {\color{red}25})&= 6+ {\color{blue}9}+ {\color{red}25} \\ (x-3)^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*} and \begin{align*}C\end{align*} are equal in the general second-degree equation.
Now, let's determine the type of conic section \begin{align*}20x^2-20y^2-80x+240y+320=0\end{align*} and rewrite the equation in standard form.
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*} and \begin{align*}y^2\end{align*} terms. Rewrite the equation, grouping the \begin{align*}x\end{align*} terms together, \begin{align*}y\end{align*} terms together, and moving the constant over to the other side. Then, pull out the GCF of each set of terms.
\begin{align*}20x^2-20y^2-90x+240y+320 &= 0 \\ 20x^2-80x-20y^2+240y &= -320 \\ 20(x^2-4x)-20(y^2-12y)&= -320\end{align*}
Now, complete the square for the \begin{align*}x\end{align*} and \begin{align*}y\end{align*} terms. When determining what constant will “complete the square” for each grouping, don’t forget to multiply the constant by the number outside the parenthesis before adding it to the other side.
\begin{align*}20(x^2-4x)-20(y^2-12y) &=\text{-}320 \\ 20(x^2-4x+ {\color{blue}4})-20(y^2-12y+ {\color{red}36}) &=\text{-}320+ {\color{blue}80}- {\color{red}720} \\ \frac{20(x-2)^2}{\text{-}960} - \frac{20(y-6)^2}{\text{-}960} &= \frac{\text{-}960}{\text{-}960} \\ \text{-}\frac{(x-2)^2}{48} + \frac{(y-6)^2}{48} &=1\\ \frac{(y-6)^2}{48}- \frac{(x-2)^2}{48} &=1\end{align*}
We now see that this conic is a hyperbola. Going back the original equation, \begin{align*}C\end{align*} is negative. In order for a general second-degree equation to be a hyperbola, \begin{align*}A\end{align*} or \begin{align*}C\end{align*} (not both) must be negative. If \begin{align*}A\end{align*} and \begin{align*}C\end{align*} are both positive or negative and not equal, the equation represents an ellipse.
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{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2}=1\end{align*}. The center is \begin{align*}(-3, 6)\end{align*}, the major axis is 14 units long, making \begin{align*}a=7\end{align*}, and the minor axis is 6 units long, making \begin{align*}b=3\end{align*}. The equation is therefore \begin{align*}\frac{(x-(-3))^2}{7^2} + \frac{(y-6)^2}{3^2}=1\end{align*} or \begin{align*}\frac{(x+3)^2}{49} + \frac{(y-6)^2}{4}=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^2-4x = 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^2-4x+ {\color{blue}4}) = 8y -12 + {\color{blue}4}\\ (x-2)^2 = 8y-8\end{align*}
Finally, factor out the LCD from the right side of the equation.
\begin{align*}(x-2)^2 = 8(y-1)\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+18x-135=0\end{align*}
Complete the square. Ellipse.
\begin{align*}9x^2+16y^2+18x-135 &=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^2-3x-8y+10=0\end{align*}
Complete the square. Parabola.
\begin{align*}y^2-3x-8y+10 &=0 \\ y^2-8y-3x &=-10 \\ y^2-8y+ {\color{red}16} &=3x-10+{\color{red}16} \\ (y-4)^2 &=3x+6 \\ (y-4)^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^2-6x+9y-14=0\end{align*}
- \begin{align*}6x^2+12x-y+15=0\end{align*}
- \begin{align*}x^2+2y^2+4x+2y-27=0\end{align*}
- \begin{align*}x^2-y^2+3x-2y-43=0\end{align*}
- \begin{align*}y^2-8x-6y+49=0\end{align*}
- \begin{align*}-64x^2+225y^2-256x-3150y-3631=0\end{align*}
Answers for Review Problems
To see the Review answers, open this PDF file and look for section 10.10.