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!

### Watch This

Khan Academy: Identifying Conics 1

### Guidance

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

Ellipses: \begin{align*}\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\end{align*}, where \begin{align*}a\end{align*} and \begin{align*}b\end{align*} are any positive numbers (the circle is the specific case when \begin{align*}a = b\end{align*}).

Parabolas: \begin{align*}y = ax^2\end{align*} or \begin{align*}x = ay^2\end{align*} where \begin{align*}a\end{align*} is any non-zero number.

Hyperbolas: \begin{align*}\frac{x^2}{a^2} - \frac{y^2}{b^2} =1\end{align*} or \begin{align*}\frac{y^2}{a^2} - \frac{x^2}{b^2} =1\end{align*}, where \begin{align*}a\end{align*} and \begin{align*}b\end{align*} are any positive numbers.

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:

\begin{align*}2x^2 + 5y^2 - 3y + 4 & = 0 \\ x^2 - 3y^2 + x - y + 3 & = 0 \\ 10x^2 - y - 5 & = 0 \\ xy & = 2\end{align*}

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 \begin{align*}-3y\end{align*} term and we will see that we have an ellipse. In the second equation we can complete the square twice to remove both the \begin{align*}x\end{align*} and \begin{align*}-y\end{align*} terms and we will have a hyperbola. This is a hyperbola, not an ellipse, because the coefficient of the \begin{align*}x^2\end{align*} and \begin{align*}y^2\end{align*} terms have opposite signs.

The third equation is a parabola since there is an \begin{align*}x^2\end{align*} term and \begin{align*}y\end{align*} term but not a \begin{align*}y^2\end{align*} term. Do you see how you can solve for \begin{align*}y\end{align*}, putting the equation in the standard form for a vertically oriented parabola?

But what about the fourth equation? Like the others, it is a degree-2 polynomial, since the exponents of the \begin{align*}x\end{align*} and \begin{align*}y\end{align*} term sum to 2. But the fourth equation looks nothing like any of the forms for conic sections that we’ve examined so far. Nonetheless, as we saw in the last section, \begin{align*}xy =2\end{align*} appears to be a hyperbola with foci (2,2) and (-2,-2). The reason it doesn’t fit either of the standard forms for hyperbolas is because it is *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?

\begin{align*}Ax^2 + By^2 + C xy + Dx + Ey + F=0\end{align*}

When \begin{align*}C=0\end{align*} we have already discussed how to determine which conic section the equation refers to. In summary, if \begin{align*}A\end{align*} and \begin{align*}B\end{align*} are both positive, the conic section is an ellipse. This is also true of \begin{align*}A\end{align*} and \begin{align*}B\end{align*} are both negative, as the entire equation can be multiplied by -1 without changing the solution set. If \begin{align*}A\end{align*} and \begin{align*}B\end{align*} differ in sign, the equation is a hyperbola, and if \begin{align*}A\end{align*} or \begin{align*}B\end{align*} equals zero the equation is a parabola.

There are a few new, more general, rules I will show you that give more information about the case when \begin{align*}C \neq 0\end{align*} and hence the conic section needs to be rotated to achieve horizontal or vertical orientation.

If \begin{align*}C^2 < 4AB\end{align*}, the equation is an ellipse (note when \begin{align*}C=0\end{align*} this holds whenever \begin{align*}A\end{align*} and \begin{align*}B\end{align*} are the same sign, which is consistent with our simpler rule stated above.)

If \begin{align*}C^2 > 4AB\end{align*}, the equation is an hyperbola (note when \begin{align*}C=0\end{align*} this holds whenever \begin{align*}A\end{align*} and \begin{align*}B\end{align*} are the opposite sign, which is consistent with our simpler rule stated above.)

If \begin{align*}C^2 = 4AB\end{align*}, the equation is a parabola (note when \begin{align*}C=0\end{align*}, either \begin{align*}A\end{align*} or \begin{align*}B\end{align*} equals zero, which is consistent with our simpler rule stated above.)

*Rotating Conics*

Suppose we have an equation of degree two polynomials, such as the \begin{align*}xy = 2\end{align*} example discussed above. In order to put it into the more recognizable form of a ellipse, parabola, or a hyperbola, we need to rotate it in such a way so that rotated version has no \begin{align*}xy\end{align*} term. So we need to find an appropriate angle \begin{align*}\theta\end{align*} such that changing the \begin{align*}x-\end{align*}coordinate to \begin{align*}x' \cos(\theta)-y' \sin (\theta)\end{align*} and the \begin{align*}y-\end{align*}coordinate to \begin{align*}x' \sin (\theta) + y' \cos (\theta) \end{align*} results in an equation with no \begin{align*}xy\end{align*} term.

#### Example A

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

*Solution:*

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

#### Example B

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

*Solution:*

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

*4ab*(400 > 12).

#### Example C

Which conic section is described by the equation?

- \begin{align*}4x^3 - 2x^2 +37 = 0\end{align*}

*Solution:*

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

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### Guided Practice

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:

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

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

2. State what type of conic section is represented by the following equation: \begin{align*}3x^2 - 5y^2 + 23 = 0\end{align*}

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

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

*Answers*

1. To identify \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*}

2. To identify \begin{align*}3x^2 - 5y^2 + 23 = 0:\end{align*}

- The x and y are both squared: not a parabola.
- The \begin{align*}x^2\end{align*} and \begin{align*}y^2\end{align*} coefficients have different signs: a hyperbola.
- \begin{align*}3x^2 - 5y^2 = -23 \to \frac{x^2}{5} - \frac{y^2}{3} = \frac{-23}{15}\end{align*}
- Since the \begin{align*}y^2\end{align*} term is negative, it has a vertical transverse
- The hyperbola's asymptotes are \begin{align*}y = \pm\frac{\sqrt{3}x}{\sqrt{5}}\end{align*}

3. To identify \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

4. To identify \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

### Explore More

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*}

### Answers for Explore More Problems

To view the Explore More answers, open this PDF file and look for section 6.9.