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# General Forms of Conic Sections

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General Conic Equation

You and your friends are playing the game Name the Conic Section. You draw a card with the equation $x^2 - 4x - 8y + 12 = 0$ . What type of conic section is represented by this equation?

### Guidance

The equation of any conic section can be written in the form $Ax^2+Bxy+Cy^2+Dx+Ey+F=0$ , which is the general second-degree equation in terms of $x$ and $y$ . For all the conic sections we studied in this chapter, $B=0$ 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 $(h, k)$

Horizontal Axis Vertical Axis
Circle $(x-h)^2+(y-k)^2=r^2$
Parabola $(y-k)^2=4p(x-h)$ $(x-h)^2=4p(y-k)$
Ellipse $\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2}=1$ $\frac{(x-h)^2}{b^2} + \frac{(y-k)^2}{a^2}=1$
Hyperbola $\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2}=1$ $\frac{(y-k)^2}{a^2} + \frac{(x-h)^2}{b^2}=1$

#### Example A

Determine the type of conic section $x^2+y^2-6x+10y-6=0$ is and rewrite the equation in standard form.

Solution: Start by rewriting the equation with the $x$ terms and $y$ terms next to each other and moving the constant to the other side of the equation.

$x^2+y^2-6x+10y-6=0 \\(x^2-6x)+(y^2+10y)=6$

Now, complete the square for both the $x$ and $y$ terms. To complete the square, you need to add $\left(\frac{b}{2}\right)^2$ to both sides of the equation.

$(x^2-6x+ {\color{blue}9}) + (y^2+10y+ {\color{red}25})&= 6+ {\color{blue}9}+ {\color{red}25} \\(x-3)^2+(y+5)^2 &= 40$

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 $A$ and $C$ are equal in the general second-degree equation.

#### Example B

Determine the type of conic section $25x^2-20y^2-100x+240y+320=0$ is and rewrite the equation in standard form.

Solution: Using the logic from the previous example, we can conclude that this conic is not a circle. It is also not a parabola because it has both the $x^2$ and $y^2$ terms. Rewrite the equation, grouping the $x$ terms together, $y$ terms together, and moving the constant over to the other side. Then, pull out the GCF of each set of terms.

$25x^2-20y^2-100x+240y+320 &= 0 \\25x^2-100x-20y^2+240y &= -320 \\25(x^2-4x)-20(y^2-12y)&= -320$

Now, complete the square for the $x$ and $y$ 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.

$25(x^2-4x)-20(y^2-12y) &=-320 \\25(x^2-4x+ {\color{blue}4})-20(y^2-12y+ {\color{red}36}) &=-320+ {\color{blue}100}+ {\color{red}720} \\\frac{25(x-2)^2}{500} - \frac{20(y-6)^2}{500} &= \frac{500}{500} \\\frac{(x-2)^2}{20} - \frac{(y-6)^2}{25} &=1$

We now see that this conic is a hyperbola. Going back the original equation, $C$ is negative. In order for a general second-degree equation to be a hyperbola, $A$ or $C$ (not both) must be negative. If $A$ and $C$ are both positive or negative and not equal, the equation represents an ellipse.

#### Example C

Write the equation of the conic below.

Solution: Just by looking at the graph, we know this is a horizontal ellipse. The standard equation for this ellipse is $\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2}=1$ . The center is $(-3, 6)$ , the major axis is 14 units long, making $a=7$ , and the minor axis is 6 units long, making $b=3$ . The equation is therefore $\frac{(x-(-3))^2}{7^2} + \frac{(y-6)^2}{3^2}=1$ or $\frac{(x+3)^2}{49} + \frac{(y-6)^2}{4}=1$ .

Intro Problem Revisit Start by rewriting the equation with the $x$ terms and $y$ terms next to each other.

$x^2-4x = 8y -12$

Now, complete the square for the $x$ terms. To complete the square, we need to add 4 to both sides of the equation.

$(x^2-4x+ {\color{blue}4}) = 8y -12 + {\color{blue}4}\\(x-2)^2 = 8y-8$

Finally, factor out the LCD from the right side of the equation.

$(x-2)^2 = 8(y-1)$

By looking at the standard forms above, we can see that this is a parabola.

### Guided Practice

Determine the conic and rewrite each equation in standard form.

1. $9x^2+16y^2+18x-135=0$

2. $y^2-3x-8y+10=0$

3. Write the equation of the conic below.

1. Complete the square. Ellipse.

$9x^2+16y^2+18x-135 &=0 \\9x^2+18x+16y^2 &=135 \\9(x^2+2x+ {\color{red}1})+16y^2 &=135+ {\color{red}9} \\9(x+1)^2+16y^2 &=144 \\\frac{(x+1)^2}{16} + \frac{y^2}{9} &=1$

2. Complete the square. Parabola.

$y^2-3x-8y+10 &=0 \\y^2-8y-3x &=-10 \\y^2-8y+ {\color{red}16} &=3x-10+{\color{red}16} \\(y-4)^2 &=3x+6 \\(y-4)^2 &=3(x+2)$

3. This is a circle because the distance around the center is the same. The center is $(0, -4)$ and the radius is 5. The equation is $x^2+(y+4)^2=25$ .

### Vocabulary

General Second-Degree Equation
$Ax^2+Bxy+Cy^2+Dx+Ey+F=0$ . For the conic sections in this chapter, $B=0$ .

### Practice

1. 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.

1. $3x^2+3y^2-6x+9y-14=0$
2. $6x^2+12x-y+15=0$
3. $x^2+2y^2+4x+2y-27=0$
4. $x^2-y^2+3x-2y-43=0$
5. $y^2-8x-6y+49=0$
6. $-64x^2+225y^2-256x-3150y-3631=0$