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/
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