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# Rotate The Axes

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Rotate the axes so that the new equation contains no xy-term. Analyze and graph the new equation.

\(x^2+4xy+y^2-3=0\)

\(x^2-4xy+y^2-3=0\)

Jun 17, 2019

#1
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Here is the first equation: I'll leave you to use the same technique on the second equation.

Jun 17, 2019
#2
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x^2 - 4xy + y^2 - 3  = 0

We have the form  Ax^2 + Bxy  + Cy^2 + F = 0

The  angle of rotation is given by

cot (2θ)  =  A - C       =      1  - 1

_____             _____  =      0

B                      -4

arccot (0)  = 2θ

pi/2  = 2θ

pi/4  = θ =  45°

Let  x  = x'cos(45) - y'sin(45)   =  [1/√2] [x' - y']

Let y  = x'sin(45) + y'cos(45) = [ 1/√2] [ x' +y']

Sub this into  x^2 - 4xy + y^2 - 3  = 0

( [1/√2] [x' - y'])^2  - 4  [1/√2] [x' - y'][ 1/√2] [ x' +y'] + ([ 1/√2] [ x' +y'])^2 - 3  = 0

(1/2)[x'^2 - 2x'y' + y'^2] - 4(1/2)[x'^2 -y'^2] +(1/2)[x'^2 + 2x'y' + y'^2] - 3 = 0

x'^2 + y'^2 - 2x'^2 + 2y'^2  - 3  = 0

3y'^2 - x'^2  =  3

y'^2 -  x'^2/3  = 1

So...this is the graph  of    y^2 - x^2/3  = 1    rotated 45°

See the graph and its rotation here : https://www.desmos.com/calculator/8lzzx1eflt   Jun 17, 2019
edited by CPhill  Jun 18, 2019
#3
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Thank you so much CPhill. This was actually the way my textbook wants me to do it so helps a lot.   