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This is a continuation from my previous question. I thought it would be better to start a new one since the old one was answered correctly.

The equation in question is: $3x^2 - 2xy - 10x +3y^2 -2y + 8=0$:

We introduce the new coordinates for $x,y$:

$$x = \cos \theta X - \sin \theta Y$$

$$y = \cos \theta Y + \sin \theta X $$

We find that $\theta = \frac{\pi}{4}$ eliminates the $xy$ term. This gives us $\cos \theta = \frac{1}{\sqrt{2}}$, $\sin \theta = \frac{1}{\sqrt{2}}$.


Now we want to find out what kind of curve $3x^2 - 2xy - 10x +3y^2 -2y + 8=0$ is.

I.E we want to complete the square and find out what kind of properties it has. What is the best method to continue from here?

One attempt is to use the general development of the $Ax^2 + Bxy +Cy^2$ with our new coordinates for $x$ and $y$. Looking at our equation above we have $A=3, B= -2, C=3$.

This gives us: $$X^2(A\cos^2 \theta + B\cos \theta \sin \theta + Csin^2 \theta)+$$

$$XY(-2A\cos \theta \sin \theta + B(\cos^2 \theta - \sin^2 \theta) + 2C \cos \theta \sin \theta) +$$ $$Y^2(A\sin^2 \theta - B\sin \theta \cos \theta + Ccos^2 \theta)$$

The coefficients for the $XY$ term will be $0$ so we can ignore that one. With the values for $A,B,C$ and $\sin \theta =\cos \theta = \frac{1}{\sqrt{2}}$ we get:

$$X^2(3\frac{1}{2} + (-2)\frac{1}{2} + 3\frac{1}{2})+ Y^2(3\frac{1}{2} - (-2)\frac{1}{2} + 3\frac{1}{2}) = 2X^2 + 4Y^2$$

The original equation $3x^2 - 2xy - 10x +3y^2 -2y + 8=0$ can now be written as: $$2X^2 + 4Y^2 -10(\cos \theta X - \sin \theta Y) -2(\cos \theta Y + \sin \theta X) + 8 = 0 \rightarrow$$ $$2X^2 + 4Y^2 -\frac{10}{\sqrt{2}}X + \frac{10}{\sqrt{2}}Y -\frac{2}{\sqrt{2}}Y - \frac{2}{\sqrt{2}} X +8 = 0 \rightarrow$$ $$2X^2 + 4Y^2 -\frac{12}{\sqrt{2}}X + \frac{8}{\sqrt{2}}Y +8 = 0$$

And if I try to complete the square I get:

$$2(x- \frac{3}{\sqrt{2}})^2+(y+2\sqrt{2})^2 = 9$$


The final result in my textbook is:

$$2(X-1)^2 + (Y-3)^2 = 3$$ where $X = x-y$ and $Y = x+y$

But I am stuck and cannot understand how to get to the result in my textbook. I would love some help (if you have the time to do the actual final calculations, that would be much appreciated!)

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    $\begingroup$ I suggest you take their answer and expand, using the defined $X=x-y, Y = x+y.$ This is not a rotation as such, but it is linear conformal, meaning, in this case, a stretched or shrunk but otherwise faithful image. That is, if you carefully draw $2(X-1)^2 + (Y-3)^2 = 3,$ the proportions will be right. $\endgroup$
    – Will Jagy
    Jun 19, 2014 at 20:18
  • $\begingroup$ Thank you Will for your comment. In the Question it said "Find a suitable rotation that eliminates the mixed term in: $3x^2 - 2xy - 10x +3y^2 -2y + 8$ and use the result to sketch $3x^2 - 2xy - 10x +3y^2 -2y + 8=0$." I don't see how the rotation helped me to find the answer they were looking for? What do you think? $\endgroup$ Jun 19, 2014 at 20:33
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    $\begingroup$ I think you should actually draw the thing by the best means available to you, also draw their version, and compare. $\endgroup$
    – Will Jagy
    Jun 19, 2014 at 20:44

1 Answer 1

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These are the steps. Show us your workings for as far as you've gotten:

  • Plug $x=(X-Y)/\sqrt{2}$ and $y=(X+Y)/\sqrt{2}$ into the equation.
  • Rearrange into the form $Ax^2+Bx^2+Cx+Dy+E=1$.
  • Complete the square on the LHS to put it in the form of the equation of an ellipse.
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  • $\begingroup$ Hi! I have updated my questions with my calculations, please have a look and see if you can please finish the final steps. $\endgroup$ Jun 19, 2014 at 16:00
  • $\begingroup$ Divide your equation by 2 and add 1 to both sides. You want it in the form: $(Ax+B)^2+(Cy+D)^2=1$. So expand this to get $A^2x^2+2ABx+B^2+C^2y^2+2CDy+D^2=1$. Now compare all of the coefficients to the ones you have in order to solve for $A,...,D$. I'm too lazy to do it myself :) $\endgroup$
    – lemon
    Jun 19, 2014 at 16:18
  • $\begingroup$ Thank you for your comment, please see my updated question regarding the final result of the question. Can you please give me some hints to how to get to the final result? $\endgroup$ Jun 19, 2014 at 18:22

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