Solving Conic Equations

Date: 05/31/99 at 14:55:32
From: Leah Whitaker
Subject: Ellipses, parabolas, and hyperbolas

I don't understand conic equations or the steps for solving the 
quadratic and putting it in the standard hyperbola, ellipse, or 
parabola form. If asked to sketch one of the above listed, I can't 
because I can't look at a problem and see the center or vertex....

Date: 06/01/99 at 09:33:56
From: Doctor Rob
Subject: Re: Ellipses, parabolas, and hyperbolas

Thanks for writing to Ask Dr. Math!

Every conic section has an equation in rectangular coordinates of 
total degree 2. In other words, you are dealing with a quadratic 
equation in the two variables x and y.

If a term involving x*y is present, then the axes of the curve are 
oblique to the coordinate axes. If it is not present, then the axes
are parallel to the coordinate axes. The former case is more 
difficult, and I won't discuss it here.

The trick to writing the equation in "standard form" is to complete 
the square on each variable which appears as a square in the equation. 
If there is just one such variable, you have a parabola, and you 
should isolate the square on one side of the equation. If there are 
two such variables, you have either an ellipse or an hyperbola, and 
you should put both squares on the same side of the equation, the 
constant on the other side, and divide by the absolute value of that 
constant. If the resulting equation has both squares with the same 
sign, you have an ellipse. If the squares have opposite signs, you 
have an hyperbola.

The standard forms are:

   Parabola:  4*p*(y-k) = (x-h)^2  (vertical axis), or
              4*p*(x-h) = (y-k)^2  (horizontal axis),
              vertex (h,k), latus rectum 4*p;

   Ellipse:   (x-h)^2/a^2 + (y-k)^2/b^2 = 1,
              center (h,k), semi-axes a, b;

   Hyperbola: (x-h)^2/a^2 - (y-k)^2/b^2 = 1, or
              (y-k)^2/b^2 - (x-h)^2/a^2 = 1,
              center (h,k), semi-axes a, b.


Example:
                        3*x^2 - 5*y^2 + 11*x + 9*y - 76 = 0,

                     3*(x^2+[11/3]*x) - 5*(y^2-[9/5]*y) = 76,

   3*(x^2+[11/3]*x+[11/6]^2) - 5*(y^2-[9/5]*y+[9/10]^2) =
                                        76 + 3*(11/6)^2 - 5*(9/10)^2,

                            3*(x+11/6)^2 - 5*(y-9/10)^2 = 2461/30,

             (x+11/6)^2/(10/2461) - (y-9/10)^2/(6/2461) = 1.

This is a hyperbola with center (-11/6, 9/10), and semi-axes
sqrt(10/2461) = 0.0637... and sqrt(6/2461) = 0.0494....

If you need help with completing the square, I suggest going to our 
Dr. Math archives and searching for   completing square  . 
There are quite a few articles there on how to do that.

I hope this helps.

- Doctor Rob, The Math Forum
  http://mathforum.org/dr.math/