# Show that $V(4xy+8xz+4yz-3y^2) \equiv V(x^2+y^2-z^2)$.

Definition An affine map from $$F^n$$ to $$F^m$$ is a function $$f : F^n \to F^m$$ given by $$f(x)=Ax+b$$ for some $$m \times n$$ matrix $$A$$ with entries in $$F$$ and some vector $$b \in F^m$$.

Definition An algebraic set $$X \subseteq F^n$$ is affinely equivalent to an algebraic set $$Y \subseteq F^m$$ if there is a bijective affine map $$f: X \to Y$$ whose inverse is also an affine map $$Y \to X$$.

Question Show that $$X=V(4xy+8xz+4yz-3y^2)$$ is affinely equivalent to $$Y=V(x^2+y^2-z^2)$$ in $$\mathbb{R}^3$$.

My Attempt: I know the key point is to express the relation $$4xy+8xz+4yz-3y^2$$ into the form $$u^2+v^2-w^2$$. This question should just be a question of mathematical tricks. I have tried to expand the relation $$(ax+by)^2+(cx+dz)^2-(ey+fz)^2$$ then match the coefficients with $$4xy+8xz+4yz-3y^2$$. But I failed.

Then I tried to expand $$(ax+by+cz)^2+(dx+ey+fz)^2-(gx+hy+iz)^2$$ and match the coefficients with $$4xy+8xz+4yz-3y^2$$. But still failed.

What should I do? This should be a question about tricks but I don't know the trick to rewrite the expression $$4xy+8xz+4yz-3y^2$$. I have been stuck on this question for two days...Can anyone help me?

Hint: The rank-3 real quadratic form $$4xy+8xz+4yz-3y^2$$ has signature $$-1$$, not $$+1$$, so match it with $$-u^2-v^2+w^2$$ instead.