<s>[INST] Find the standard form of the conic section $x^2-3x+4xy+y^2+21y-15=0$ Find the standard form of the conic section $x^2-3x+4xy+y^2+21y-15=0$.
I understand the approach in trying to solve these problems. But the $4xy$ is confusing me. I am not sure of where to start on this one.
UPDATE:
i have found a way of doing this with reference to Canonical form of conic section
Solution:
$x^2-3x+4xy+y^2+21y-15=(x+2y)^2-3y^2-3x+21y$
Let: 
$X = x+2y$
$Y=y$
$x = X - 2Y$
Now we substitute:
$X^2-3Y^2-3X+27Y-15=0$
so it is a hyperbola?
Please check.
 [/INST] Consider the second degree terms: $x^2 + 4xy + y^2$. In general, if the second degree term is $Ax^2 + Bxy + Cy^2$, then the discriminant is $D = B^2 - 4AC$. If the conic section is non-degenerate, then
\begin{align}
D > 0 & \implies \text{Hyperbola}\\
D = 0 & \implies \text{Parabola}\\
D < 0 & \implies \text{Ellipse}
\end{align}
Hence, in your case, it can be a hyperbola or a pair of straight lines.
Now factorize the second degree term as follows.
$$x^2 + 4xy + y^2 = \underbrace{(x+(2+\sqrt{3}) y)}_{X}\underbrace{(x+(2-\sqrt{3}) y)}_{Y}$$
$$x^2 + 4xy + y^2 -3x + 21 y - 15 = XY + aX + bY - 15 =0$$
Hence, we get that $a+ b= -3$ and $(2+\sqrt{3})a+ (2-\sqrt{3})b= 21$.
This gives us $a = \dfrac32 (-1 + 3 \sqrt{3})$ and $b = \dfrac32(-1 - 3 \sqrt{3})$.
Hence, $$x^2 + 4xy + y^2 -3x + 21 y - 15 = XY + aX + bY - 15 = (X+b)(Y+a) - ab - 15 = 0$$
Hence, $$(X+b)(Y+a) = 15 - \dfrac{117}2 = -\dfrac{87}2$$
in the canonical hyperbola form.
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