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
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.