Identifying Conics Solutions

• When given a    graph     of a conic, identify the    shape     to determine if it is a parabola, circle, ellipse, or hyperbola.

• When given an    equation     of a conic that is not in standard form, complete the square to determine the type of conic.

• Writing equations in standard form (or vertex form for parabolas) is important for identifying:

  • the    type     of conic section that is given.

  • individual    elements     of each type of conic.

• Conic sections can be identified when written in the form $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$ by identifying the distinguishing characteristics of each conic section using the coefficients    A, B, and C    . 

  • A, B, and C     are not     all equal to zero.

  • A–F are    constants    .

• Using the sample sentence in the table can help you identify conic sections:

When…	the conic section is a(n) ________.
A or C is zero	parabola
A = C and are non-zero	circle
$A\ne C$ and are non-zero and have the same sign	ellipse
A and C are non-zero and have opposite signs	hyperbola

Example 5

Name the conic sections. Explain.
$\begin{array}{ccc}P: x^2+36y^2–72x=180 & Q: 2x^2–8y^2+8x+32=16y & R: y^2–9x+18y+81=0\end{array}$

Plan

Identify A and C

Determine the conic

P	$A=1, C=36$ Since A and C are   non-zero and have the same sign  , equation P represents   ellipse  .
Q	$\mathit{A}\mathbf{=}\mathbf{2}\mathbf{,}\mathbf{ }\mathit{C}\mathbf{=}\mathbf{–}\mathbf{8}$ Since A and C are   non-zero and have opposite signs   equation Q represents   hyperbola  .
R	$\mathit{A}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{ }\mathit{C}\mathbf{=}\mathbf{1}$ Since $\mathit{A}\mathbf{=}\mathbf{0}$, equation R represents a parabola

Example 6

Determine the type of conic. Then name the domain and range.


$x^2+y^2+9y+1.5=5x$

$A=1, C=1$


Since $\underline{\mathbf{ }\mathbf{ }\mathbf{ }\mathit{A}\mathbf{=}\mathit{C}\mathbf{ }\mathbf{ }\mathbf{ }}$, and are non-zero, the equation is    a circle   . 


$$\begin{gathered} x^2–5x+y^2+9y=–1.5 \\ \left(x^2–5x+\boxed{{\left(–\frac{5}{2}\right)}^2}\right)+\left(y^2+9y+\boxed{{\left(\frac{9}{2}\right)}^2}\right)=–1.5+{\left(–\frac{5}{2}\right)}^2+{\left(\frac{9}{2}\right)}^2 \\ {\left(x–\frac{5}{2}\right)}^2+{\left(y+\frac{9}{2}\right)}^2=–\frac{6}{4}+\frac{25}{4}+\frac{81}{4} \\ {\left(x–\frac{5}{2}\right)}^2+{\left(y+\frac{9}{2}\right)}^2=25 \end{gathered}$$

Note

Because more information is needed in addition to the type of conic, completing the square is necessary.

$$\begin{gathered} \mathrm{Center} \left(2.5, –4.5\right), r=5 \\ \mathrm{D}\mathrm{omain}=2.5\pm 5, \mathrm{R}\mathrm{ange}=–4.5\pm 5 \\ \mathrm{D}\mathrm{omain} \left[–3.5, 7.5\right] \\ \mathrm{R}\mathrm{ange} \left[–9.5, 0.5\right] \end{gathered}$$