Last modified on 25 February 2011, at 23:32

Conic Sections/Identifying Conics

Identifying ConicsEdit

Conics are reasonably easy to graph if they are given in their standard form. Unfortunately, this is often not the case. Conics can be represented by polynomials of the form: Ax^2 + Bxy +Cy^2 + Dx + Ey + F = 0, and to graph these it is necessary to convert them back to their standard form. Before you do this, though, the formula: B^2 -4AC can be used to determine the type of conic from the original equation before you start graphing:

  • B^2 -4AC = 0: Parabola
  • B^2 -4AC < 0: Ellipse
  • B^2 -4AC > 0: Hyperbola

Firstly, if B is not zero then the graph represents a rotated conic. Follow the instructions given on Rotation of Axes to determine how to convert this to a non-rotated form. Then convert this to the standard form of a conic as detailed later in this section, and graph it on a set of axes at the appropriate angle to the x- and y- axes.

If B is equal to zero, then use completing the square and algebraic manipulation to obtain standard forms of the conics.

ExamplesEdit

Example 1: (Ellipse) \frac{1}{9}x^2+\frac{1}{25}y^2-\frac{2}{3}x-\frac{8}{25}y+\frac{16}{25}=0Edit

It is first necessary to complete the square for the x and y terms. First we will start with x. While it appears that in this case b is -\frac{2}{3}, you must remember that the coefficient of x^2 must be equal to 1. Therefore: \frac{1}{9}x^2-\frac{2}{3}x
=\frac{1}{9}[x^2-6x]
Now, b = -6. (\frac{b}{2})^2=(\frac{-6}{2})^2=9. Therefore:
=\frac{1}{9}[x^2-6x+9-9]
=\frac{1}{9}[(x-3)^2-9]
=\frac{1}{9}(x-3)^2-1
Doing the same for the y: \frac{1}{25}y^2-\frac{8}{25}y
=\frac{1}{25}[y^2-8y]
Now, b = -8. (\frac{b}{2})^2=(\frac{-8}{2})^2=16. Therefore:
=\frac{1}{25}[y^2-8y+16-16]
=\frac{1}{25}[(y-4)^2-16]
=\frac{1}{25}(y-4)^2-\frac{16}{25}
These can then be placed back into the original equation: (\frac{1}{9}(x-3)^2-1)+(\frac{1}{25}(y-4)^2-\frac{16}{25})+\frac{16}{25}=0
Which simplifies to \frac{(x-3)^2}{9}+\frac{(y-4)^2}{25}=1 Which is the standard form for an ellipse. To check our answer, we can use the formula B^2 -4AC getting the values from the original equation. Substituting these in gives: (0)^2 -4(\frac{1}{9})(\frac{1}{25})=-\frac{4}{225} which is less than zero. Therefore the conic is an ellipse, confirming our previous answer.

Example 2: CircleEdit

Example 3: ParabolaEdit

Example 4: HyperbolaEdit

Example 5: (Rotated Ellipse) 6x^2 + \sqrt{3}xy+7y^2-36=0Edit

The fact that it has an xy term means that this is a rotated conic. The angle that it has been rotated is given by: tan 2\theta = \frac{B}{A-C}=\frac{(\sqrt{3})}{(6)-(7)}
Therefore 2\theta=atan \frac{(\sqrt{3})}{(6)-(7)}
2\theta=atan -\sqrt{3}
2\theta=\frac{-\pi}{3}
\theta=\frac{-\pi}{6}
This tells you that the conic has been rotated -\frac{\pi}{6} radians, or -30°, anticlockwise, which is the same as \frac{\pi}{6} radians or 30° clockwise about the origin.
You can then make the following substitutions to obtain the polynomial for an identical, non-rotated conic:

  • x = X cos \theta - Y sin \theta
  • y = X sin \theta + Y cos \theta

In this case, \theta is -\frac{\pi}{6}. Make sure you do not confuse the sign as this will make further calculations invalid. the value for the angle used is always the amount anti-clockwise. Therefore, substituting \theta = -\frac{\pi}{6} into the previous will give you:

  • x = X cos \frac{-\pi}{6} - Y sin \frac{-\pi}{6}
  • y = X sin \frac{-\pi}{6} + Y cos \frac{-\pi}{6}

Which simplifies to

  • x = \frac{\sqrt{3}}{2}X - \frac{1}{2}Y
  • y = -\frac{1}{2}X + \frac{\sqrt{3}}{2}Y

When these are substituted into the rotated equation, you get:
6(\frac{\sqrt{3}}{2}X - \frac{1}{2}Y)^2 + \sqrt{3}(\frac{\sqrt{3}}{2}X - \frac{1}{2}Y)(-\frac{1}{2}X + \frac{\sqrt{3}}{2}Y)+7(-\frac{1}{2}X + \frac{\sqrt{3}}{2}Y)^2-36=0
Expanding the brackets gives:
6(\frac{3}{4}X^2+\frac{\sqrt{3}}{2}XY+\frac{1}{4}Y^2)+\sqrt{3}(-\frac{\sqrt{3}}{4}X^2+\frac{1}{2}XY+\frac{\sqrt{3}}{4}Y)+7(\frac{1}{4}X^2-\frac{\sqrt{3}}{2}XY+\frac{3}{4}Y^2)-36=0
When you expand the brackets and collect like terms, you end up with:
(6\times\frac{3}{4}+\sqrt{3}\times\frac{-\sqrt{3}}{4}+7\times\frac{1}{4})X^2+(6\times\frac{\sqrt{3}}{2}+\sqrt{3}\times\frac{1}{2}+7\times\frac{-\sqrt{3}}{2})XY+(6\times\frac{1}{4}+\sqrt{3}\times\frac{\sqrt{3}}{4}+7\times\frac{3}{4})Y^2-36=0
Which simplifies to:
\frac{11}{2}X^2+0XY+\frac{15}{2}Y^2-36=0
The fact that there is now no XY term means that this is an equation for a non-rotated conic. Now all that remains is a little algebraic manipulation to get it into its standard form:
\frac{11}{2}X^2+\frac{15}{2}Y^2=36
\frac{X^2}{\frac{2}{11}}+\frac{Y^2}{\frac{2}{15}}=36
\frac{X^2}{\frac{2}{11}\times36}+\frac{Y^2}{\frac{2}{15}\times36}=1
\frac{X^2}{\frac{72}{11}}+\frac{Y^2}{\frac{24}{5}}=1
Which is the standard form of an ellipse which you can use to find the axis lengths, eccentricity, foci and anything else. Remember though that this must be graphed at the angle found earlier of \frac{-\pi}{6} to the x-axis. You can use the same formulae from above to find the rotated location of points from the non-rotated ellipse:
The point (x,y) when rotated by an angle of \theta about the origin becomes (X,Y) where:

  • X = x cos \theta + y sin \theta
  • Y = -x sin \theta + y cos \theta