$\newcommand{\bs}{\boldsymbol{\sigma}} \newcommand{\bg}{\boldsymbol{\gamma}}\newcommand{\bp}{\textbf{p}}$ Let $\bs(u,v)=(1-v)\bp+v\bg(u)$ be a generalized cone, where $\bg$ is unit speed curve and $\bp$ is a fixed point. In the book, the author derived the first fundamental form of the generalized cone by using that the first fundamental form does not change applying an isometry of $\mathbb{R}^3$ to the surface and the result is $v^2du^2+dv^2$. But when I am cauculating without considering isometry I am getting a rather unlikely expression. Can anyone show me how to calculate it without considering isometry?
Thanks.
$\partial_u=v \gamma'(u)$
$\partial_v=-p+\gamma(u)$
$\langle\partial_u,\partial_u \rangle=v^2\Vert\gamma'(u) \Vert^2=v^2$
$\langle\partial_u,\partial_v \rangle=v\langle \gamma'(u),-p \rangle+v\langle \gamma'(u),\gamma(u) \rangle$
$\langle\partial_v,\partial_v \rangle=\Vert p \Vert^2 -2 \langle p,\gamma(u) \rangle+ \Vert \gamma(u) \Vert^2$
Hence the first fundamental form in it's general form is given by: $$\begin{pmatrix}v^2& v\langle \gamma'(u),-p \rangle+v\langle \gamma'(u),\gamma(u) \rangle\\ v\langle \gamma'(u),-p \rangle+v\langle \gamma'(u),\gamma(u) \rangle & \Vert p \Vert^2 -2 \langle p,\gamma(u) \rangle+ \Vert \gamma(u) \Vert^2 \end{pmatrix}$$ Applying translation to $\mathbb R^3$ we can asume that the cone is centered at the origin, $p=0$. F.f.f becomes then $$\begin{pmatrix}v^2& v\langle \gamma'(u),\gamma(u) \rangle\\v\langle \gamma'(u),\gamma(u) \rangle & \Vert \gamma(u) \Vert^2 \end{pmatrix}$$
Furthermore, it's reasonable to say that $\gamma$ lies in the sphere ($\Rightarrow \Vert \gamma \Vert\equiv 1$), if not we substitute $\bar \gamma(u):=\frac {\gamma(u)}{\Vert \gamma(u) \Vert}$ in coordinate patch. (Now if such construction holds for all cones, depends on definition of generalised cone.) Above implies $\langle\gamma, \gamma' \rangle\equiv 0$. And we get $$\begin{pmatrix}v^2& 0\\0 & 1 \end{pmatrix}$$ as desired.