Graphing and Understanding Conic Equations

Date: 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
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