Graphing and Understanding Conic EquationsDate: Mon, 11 Dec 95 00:37 EST From: Anonymous Subject: Question Here's a question from an OAC Algebra and Geometry Assignment that's got me stumped: By a suitable rotation and translation, reduce the equation to the standard form of a conic, then sketch it and determine its vertices. x^2 - 4xy + y^2 + 4*Sqrt{2}*x - 2*Sqrt{2}*y + 11 = 0 I can figure out how to get it into standard form, and but I can't figure out how to sketch it or find it's vertices. Any help you could give would be great. Thanks. Greg McIlwain, Toronto, Ontario. From: Doctor Jerry Subject: Re: Question I assume that you found the equation (I'll use X and Y for the new coordinates) x^2 - 3y^2 - 2x + 6y = 11 This results from a rotation of pi/4. Now, just complete the squares to determine the translation of axes needed to reduce the new equation to standard form. x^2 - 2x + 1 -3(y^2 - 2y +1) = 11 + 1 - 3 (x-1)^2 - 3(y-1)^2 = 9 With new coordinates P and Q, with origin at (1,1) in the (X,Y) system, i.e. P=x-1 and Q=y-1: P^2 - 3Q^2 = 9 (P^2)/9 - (Q^2)/3 = 1 (P/3)^2 - (Q/Sqrt{3})^2 = 1 This is, as I'm sure you know, a hyperbola. Thanks for your question. -Doctor Jerry, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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