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Given the conic section equation

$$x^2-2\sqrt {3}xy+3y^2+y=0$$

determine its type, and find its vertex, focus, and directrix.

My try :

$$\text{discriminant} = B^2-4AC=12 - 4\cdot 1\cdot 3 = 0$$

since $\text{discriminant}=0$ , the conic section is a parabola.

Now, how can I write the above parabola in standard form so that I can get all needed information?

Thank you

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    $\begingroup$ $4((y+\frac1{16})^2+(x-\frac{\sqrt{3}}{16})^2-(\frac12 y+\frac{\sqrt{3}}{2} x -\frac18)^2)$ simplifies to your equation, meaning the focus is $(\frac{\sqrt{3}}{16},-\frac1{16})$ and the directrix is $\frac12 y+\frac{\sqrt{3}}{2}x-\frac18=0$ $\endgroup$ – Jan-Magnus Økland Jun 19 '18 at 11:20
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Notice that $$ x^2−2\sqrt3xy+3y^2=(x-\sqrt3y)^2= 4\left(x\cos{\pi\over3}-y\sin{\pi\over3}\right)^2. $$ The expression inside the last parentheses is the resulting $x$ coordinate when point $(x,y)$ is rotated by $\pi/3$ counterclockwise. Introducing then the rotated coordinates: $$ X=x\cos{\pi\over3}-y\sin{\pi\over3}, \quad Y=y\cos{\pi\over3}+x\sin{\pi\over3}, $$ your equation becomes $$ Y=-8X^2+\sqrt3X, $$ which is in standard form. You can then find focus, vertex, and so on, for this parabola and then rotate back by $-\pi/3$ to the original $(x,y)$ coordinate system.

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