How can we simplify an ellipse of form $Ax^2 + Bxy + Cy^2 = F$

Question

How can we simplify such ellipse to the form of an ellipse $$\frac{{(x - h)^2}}{{a^2}} + \frac{{(y - k)^2}}{{b^2}} = 1$$ As far as I know and could simplify was when $B$ in such second-degree equations is zero. Because then even if we have the general second-degree equation $$Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$$ We have $B = 0$ then we can complete the square and still obtain the desired equation of the ellipse in the form of $$\frac{{(x - h)^2}}{{a^2}} + \frac{{(y - k)^2}}{{b^2}} = 1$$

I am not sure how to go about doing this! As an example I have been trying to simplify $3x^2 + 2xy +3y^2 = 8$ But to no avail. Thanks for any advice!

Whether or not this will be an ellipse depends on the sign of $4AC-B^2$ (and also on $F$). This will not always be an ellipse. — Mikhail Katz, Oct 02 '23 at 15:09
Do you know anything about diagonalizing a 2 by 2 matrix? You need to be able to find the eigenvalues and eigenvectors of the 2 by 2 matrix. — Paul, Oct 02 '23 at 15:10
@Paul I do not, we have just been introduced to ellipses and all we know is this general formula and that foci can be found by $c^2 = a^2 - b^2$ and that the general second degree equation can describe them specifically when $B = 0$ — Teodoras Paura, Oct 02 '23 at 15:12
This is a rotation of axes problem. The angle of rotation $\theta$ has cotangent $(A-C)/B$. The hints in the other comments should help you proceed. Conic sections generally are handled in some modern linear algebra texts. — Chris Leary, Oct 02 '23 at 15:13
@ChrisLeary but in my given example A = 3, C = 3 so the cotangent = 0 so undefined even though this is an ellipse? — Teodoras Paura, Oct 02 '23 at 15:33
From your equation in the form $$8(\frac{(\frac{x+y}{\sqrt2})^2}{\sqrt2^2}+\frac{(\frac{x-y}{\sqrt2})^2}{2^2}-1)=0$$ you can read off the semi-major axis of $2$ and semi-minor axis of $\sqrt2.$ — Jan-Magnus Økland, Oct 02 '23 at 16:58
@Jan-MagnusØkland Yup, that's what I got with the angle rotation! Thank you:) Because, unlike the answer below states, you cannot complete the square.. — Teodoras Paura, Oct 02 '23 at 17:01
@TeodorasPaura - There is a crucial word missing in my earlier comment. It should read twice the angle of rotation ... . I learned of these problems back in the dark ages from the third edition of Thomas' Calculus. As Mikhail commented, what conic you get depends on the sign of $B^2-4AC.$ You get a parabola if it is $0$, an ellipse if negative, and a hyperbola if positive. — Chris Leary, Oct 02 '23 at 19:16
@ChrisLeary Thank you! Would you recommend working with that book? Or maybe you would have different recommendations? I am a double BSc student in math and physics. — Teodoras Paura, Oct 02 '23 at 22:16
@TeodorasPaura - That edition of Thomas has been out of print for years (I used it in 1966-1967). I don't recall seeing any calculus book treat rotation of axes after 1990 or so. Apostol discusses rotation of axes relative to conic sections in the second volume of his calculus book. It all has to do with eigenvalues of orthogonal matrices. As I said earlier, your best bet may be to check out some recent linear algebra texts in the library. I retired about seven years ago and gave away all my textbooks. Otherwise, I could probably give you specific suggestions. — Chris Leary, Oct 02 '23 at 23:38
@TeodorasPaura - You're welcome. Best of luck with your studies. — Chris Leary, Oct 03 '23 at 14:10

score 3 · Accepted Answer · answered Oct 02 '23 at 15:44

Hint.

Supposing that $3 x^2 + 2 x y + 3 y^2 - 8=0$ represents an ellipse, making the change of variables

$$ \left(\matrix{x\\ y}\right) = \left( \begin{array}{cc} \cos (\theta ) & -\sin (\theta ) \\ \sin (\theta ) & \cos (\theta ) \\ \end{array} \right)\left(\matrix{X\\ Y}\right) $$

we get

$$ \sin (2 \theta ) (X^2-Y^2)+2 X Y \cos (2 \theta )+3 X^2+3 Y^2-8=0 $$

now choosing $\theta = \frac{\pi}{4}$ to eliminate the product $XY$ we arrive at

$$ 4 X^2 + 2 Y^2 - 8 = 0 $$

How can we simplify an ellipse of form $Ax^2 + Bxy + Cy^2 = F$

1 Answers1