Proving that the cone is isometric to the plane There is a given cone, $ 3z = 4 \sqrt{x^2+y^2} $. I have to prove that this cone is isometric to the plane.
First, I parametrized the cone: $ r(u,v)=(u \cos v, u \sin v, \frac{4}{3} u) $. The coefficients of the first fundamental form are $ E = 1+\frac{16}{9}, F=0, G=u^2 $.
The next step would be finding the plane which has the same coefficients of the first fundamental form.
How do I find the right plane?
In my mind, the only right plane would be $ z=0 $. So, I took the curve from the given surface: $ \alpha(t) = (\cos t, \sin t) $. This is natural parameterization, so $t$ is natural parameter, $s(t)=t$. Next, I'm finding the plane: (u,v,0). The coefficients of the first fundamental form are $E=1, F=0, G=1$, but that's not what I was looking for.
Where did I go wrong? I suppose this is not the right way the find the plane.
Thank you.
 A: Let's build some geometric intuition first. How do you turn a sheet of paper into a cone? You cut out a circular sector and glue its boundary radii together. I'd suggest to parametrize the plane to match this intuition. Let $v$ be proportional to an angle around the origin (proportional because a full circle around the cone is less than a full circle in the plane), and $u$ proportional to the radius.
One next step would be simply starting with these assumed proportionalities, and working out the corresponding ratios without spending much more thought on them. For that I'd start with
$$r'(u,v)=(\cos(\alpha v)\beta u,\sin(\alpha v)\beta u,0)$$
Now compute the first fundamental form of this, by computing the derivatives.
$$
r'_u(u,v)=(\cos(\alpha v)\beta,\sin(\alpha v)\beta,0)
\qquad
r'_v(u,v)=(-\sin(\alpha v)\alpha\beta u,\cos(\alpha v)\alpha\beta u,0)
$$
The coefficients then read
$$
E'=\beta^2\overset!=\frac{25}9\qquad
F'=0\qquad
G'=\alpha^2\beta^2u^2\overset!=u^2
$$
and therefore the factors of proportioinality are
$$
\beta=\pm\frac53\qquad
\alpha=\pm\frac35
$$
