I have started with $$x^2-2xy+y^2+y=0$$
Using eigenvectors and eigenvectors and completing the square I came to
$$(y'-\frac{1}{\sqrt{32}})^2+\frac{1}{\sqrt{8}}x'=\frac{1}{\sqrt{32}}$$
How can I conclude which conic section it is?
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1This looks like $u^2+v=0$, and that looks like a parabola. – Angina Seng Nov 10 '17 at 07:26
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https://en.wikipedia.org/wiki/Rotation_of_axes#Rotation_of_conic_sections – lab bhattacharjee Nov 10 '17 at 07:26
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@labbhattacharjee Yes I am looking for the standard form – gbox Nov 10 '17 at 07:28
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See whether this helps you: https://www.ck12.org/calculus/classifying-conic-sections/lesson/Classifying-Conic-Sections-ALG-II/ – Nov 10 '17 at 07:33
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This answer expands on @labbhattacharjee's comment.
According to this Math SE post, you can use the equation to find the $A,B,C$ in the general form $Ax^2+Bxy+Cy^2+Dx+Ey+F$, then find the discriminant $B^2 - 4AC$.
In this case, we have $(-2)^2 - 4(1)(1)$ which equals $0$. Since $B^2 - 4AC = 0$, the equation represents a parabola. You can use this method quite easily with other equations as well.
Toby Mak
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