Kelli was reviewing for her pre-calc final on conic sections, and she was trying to memorize all of the rules for each of the shapes. She was actually doing pretty well, but was frustrated with the challenge of keeping them all straight.

Kelli found herself wishing there were a way to identify the shape early on, so she would have an idea of what she was generally sketching before she worried about specifics.

Too bad Kelli didn't know about the material in this lesson!

### General Forms of Conic Sections

Let’s examine all the equations of the conic sections we’ve studied in this chapter.

Ellipses:

Parabolas:

Hyperbolas:

All these equations have in common that they are degree-2 polynomials, meaning the highest exponent of any variable—or sum of exponents of products of variables—is two. So for example, here are some degree two polynomial equations in a more general form:

Some of these probably already look like conic sections to you. For example, in the first equation, we can complete the square to remove the

The third equation is a parabola since there is an

But what about the fourth equation? Like the others, it is a degree-2 polynomial, since the exponents of the *diagonally* oriented, rather than horizontally or vertically oriented (do you see how its two foci lie on a diagonal line, rather than a horizontal or vertical line?)

In order to see how such differently-oriented conic sections fit into our standard forms, we need to rotate them so that they are either horizontally or vertically oriented.

#### Identifying Conics

How can we look at a degree-2 polynomial equation and determine which conic section it depicts?

When

There are a few new, more general, rules I will show you that give more information about the case when

If

If

If

#### Rotating Conics

Suppose we have an equation of degree two polynomials, such as the

### Examples

#### Example 1

State what type of conic section is represented by the following equation: \begin{align*}5x^2+6y^2+2x-5y+xy=0\end{align*}.

This is an ellipse, as both *a* and *b* are positive. Also, note that *c* < *4ab*.

#### Example 2

State what type of conic section is represented by the following equation: \begin{align*}x^2+3y-20xy+20=0\end{align*}.

This is a hyperbola, since *c** ^{2}* >

*4ab*(400 > 12).

#### Example 3

Which conic section is described by the following equation: \begin{align*}4x^3 - 2x^2 +37 = 0\end{align*}?

To identify \begin{align*}4x^3 - 2x^2 +37 = 0\end{align*} we just need to look at the signs of the squared terms:

The \begin{align*}x^2\end{align*} and \begin{align*}y^2\end{align*}coefficients have different signs so it's a hyperbola.

Additionally, the \begin{align*}y^2\end{align*} term is negative, so it has a vertical transverse.

**For the following examples, use the information below:**

When identifying conic sections, the squared terms are the only factors that matter. Run the **standard form** of the equation you are questioning through the following series of tests on the squared terms:

- If only 1 variable is squared, you have a parabola, stop further testing.
- If the coefficients of the squared terms have opposite signs, you have a hyperbola, stop further testing.
- If the squared terms are multiplied by the same coefficient, you have a circle, stop testing.
- If none of the above apply, you have an ellipse.

#### Example 4

Identify and describe the conic section represented by the following equation: \begin{align*}(x - 1)^2 + (y + 1)^2 = 9\end{align*}.

Both terms are squared: not a parabola

The coefficients of \begin{align*}x^2\end{align*} and \begin{align*}y^2\end{align*} are both positive: not a hyperbola

Both squared term coefficients are the same: a circle

\begin{align*}r^2 = 9 \to r = 3\end{align*}

The center \begin{align*}(h, k)\end{align*} is \begin{align*}(1, -1)\end{align*}.

#### Example 5

State what type of conic section is identified by the following equation: \begin{align*}-8x^2 - 18y^2 + 48x - 360y - 1800 = 0\end{align*}.

\begin{align*}-8(x - 3)^2 - 18(y + 10)^2 = -72\end{align*} ..... Complete the square

\begin{align*}\frac{(x - 3)^2}{9} + \frac{(y + 10)^2}{4} = 1\end{align*} ..... Simplify

Both \begin{align*}x\end{align*} and \begin{align*}y\end{align*} are squared: not a parabola

Both coefficients are negative: not a hyperbola

The \begin{align*}x^2\end{align*} and \begin{align*}y^2\end{align*} terms do not have the same coefficients: not a circle either

This ellipse has center: (3, -10)

From the center, plot 2 up and \begin{align*}\sqrt{9}\end{align*} left and right to mark the vertices.

#### Example 6

State and describe the conic section: \begin{align*}2x^2 - 16x + 32 +3y = 6\end{align*}.

Only the \begin{align*}x\end{align*} term is squared, this is a parabola

\begin{align*}2(x-4)^2 + 3(y - 2)= 0\end{align*} ..... Factor and simplify

The vertex \begin{align*}(h, k)\end{align*} is \begin{align*}(4, 2)\end{align*}.

The \begin{align*}x\end{align*} term is squared, this is an up/down parabola

The \begin{align*}x^2\end{align*} term is positive, this parabola opens down

The parent \begin{align*}y = x^2\end{align*} has been compressed vertically by 3 and horizontally by 2, resulting in a slightly wider than standard curve.

### Review

Identify the conic that is represented by the equation.

- \begin{align*} \frac{(x - 5)^2}{4} + \frac{(y - 4)^2}{4} = 1\end{align*}
- \begin{align*} (x - 3)^2 + y - 2^2 = 1\end{align*}
- \begin{align*} x^2 - 5x - y^2 - 4y + 16 = 0\end{align*}
- \begin{align*} \frac{2y}{3} = 2(x-4)^2 + 1\end{align*}

Convert to standard form:

- \begin{align*}-8x^2 - 18y^2 + 48x - 360y - 1800 = 0\end{align*}
- \begin{align*}-12x^2 + 12y^2 - 72x + 192y + 708 = 0\end{align*}
- \begin{align*}x^2 + y^2 -8x + 10y + 32 = 0\end{align*}
- \begin{align*}9x^2 + 4y^2 - 36x + 64y + 256 = 0\end{align*}
- \begin{align*}9x^2 + y^2 + 90x - 8y + 232 = 0\end{align*}
- \begin{align*}9x^2 - 9y^2 + 162x + 162y - 81 = 0\end{align*}

Identify and graph the following:

- \begin{align*}\frac{(x + 4)^2}{9} + (y)^2 = 1\end{align*}
- \begin{align*}y = -2(x + 4)^2 - 4\end{align*}
- \begin{align*}(x + 2)^2 - (y - 2)^2 = 1\end{align*}
- \begin{align*}\frac{(x - 1)^2}{4} = \frac {(y)^2}{16} = 1 \end{align*}
- \begin{align*}x = 2(y + 2)^2 - 1\end{align*}
- \begin{align*}x = -(y - 2)^2 + 4\end{align*}
- \begin{align*}\frac{(x + 2)^2}{1} - \frac{(y - 2)^2}{1} = 1\end{align*}
- \begin{align*}\frac{(y - 1)^2}{4} - \frac{(x - 4)^2}{4} = 1 \end{align*}

### Review (Answers)

To see the Review answers, open this PDF file and look for section 6.9.