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### 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/
```
Associated Topics:
High School Equations, Graphs, Translations

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