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avatar+895 

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
avatar+28354 
+3

Here is the first equation:

 

I'll leave you to use the same technique on the second equation.

 Jun 17, 2019
 #2
avatar+106519 
+1

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

 

 

cool cool cool

 Jun 17, 2019
edited by CPhill  Jun 18, 2019
 #3
avatar+895 
+1

Thank you so much CPhill. This was actually the way my textbook wants me to do it so helps a lot.

AdamTaurus  Jun 17, 2019
 #4
avatar+106519 
0

OK....glad to help.....

 

cool cool cool

CPhill  Jun 18, 2019

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