Writing the metric of 4D de Sitter space as a 5D metric It is often said that the 4D de Sitter metric may be obtained as the induced metric on a hyperboloid embedded in 5D Minkwoski space.  The hyperboloid is
$$
-x_0^2+x_1^2+x^2_2+x_3^3+x^2_4=l^2~~.
$$
I seem to recall from my undergraduate cosmology studies that the 4D de Sitter metric can be represented as a 5D metric with the $l^2$ parameter in the fifth diagonal position.  However, I am unable to find the formula I am looking for.  For instance, the 5D Minkowski metric may be written
$$
ds^2=-dx_0^2+dx^2_1+dx^2_2+dx^2_3+dx^2_4~~.
$$
If I replace $dx_4^2$ with $l^2$, that seems to have a problem because it requires that a sum of differentials is equal to a finite number $l^2$.   So, I am not remembering something about a way to write the metric of dS$_4$ as a simpler 5D metric.
What is meant when they say that the de Sitter metric can be represented as a 5D metric with the fifth metric value being the curvature parameter?
 A: What you want is the induced metric. I am not going to solve this explicitly for you since this is a very standard homework/exam exercise. But let me sketch the identical computation for a sphere.
A 2d sphere can be embedded in 3d via
$$
x^2+y^2+z^2=R^2
$$
The flat 3d metric is
$$
ds^2(\mathbb R^3)=dx^2+dy^2+dz^2
$$
If you solve for $z$ in the equation that defines the sphere, you get
$$
z=\sqrt{R^2-x^2-y^2}
$$
i.e.,
$$
dz=\frac{-xdx-ydy}{\sqrt{R^2-x^2-y^2}}
$$
Plugging this back into the flat metric, you get
$$
ds^2(S^2)=z^*ds^2(\mathbb R^3)=dx^2+dy^2+\frac{(xdx+ydy)^2}{R^2-x^2-y^2}
$$
which is the metric of the 2d sphere. Of course, you can change coordinates if you want to obtain more traditional presentations.
--
In your case, the idea is identical. You have a formula that specifies the embedding of dS into flat 5d space. Solve for one of the coordinates and take the differential. Plug the result into the flat metric to obtain the metric induced by the embedding.
A: I use the solution of @AccidentalFourierTransform and write you the results for your case
with
$$-{x_{{0}}}^{2}+{x_{{1}}}^{2}+{x_{{2}}}^{2}+{x_{{3}}}^{2}+{x_{{4}}}^{2}=l^2\tag 1$$
and
$$ds^2=-{{\it dx}_{{0}}}^{2}+{{\it dx}_{{1}}}^{2}+{{\it dx}_{{2}}}^{2}+{{\it 
dx}_{{3}}}^{2}+{{\it dx}_{{4}}}^{2}
\tag 2$$
solve the "constrain"  equation (1)  for $~x_4~$ you obtain
$$x_4=\pm\sqrt {{x_{{0}}}^{2}-{x_{{1}}}^{2}-{x_{{2}}}^{2}-{x_{{3}}}^{2}+{l}^{2
}}$$
thus
$$dx_4=\pm{\frac {x_{{0}}{\it dx}_{{0}}-x_{{1}}{\it dx}_{{1}}-x_{{2}}{\it dx}_{
{2}}-x_{{3}}{\it dx}_{{3}}}{\sqrt {{x_{{0}}}^{2}-{x_{{1}}}^{2}-{x_{{2}
}}^{2}-{x_{{3}}}^{2}+{l}^{2}}}}
$$
from here you obtain the new line element or the metric
$$G=\pm\frac{1}{d}\,\left[ \begin {array}{cccc} {x_{{1}}}^{2}+{x_{{2}}}^{2}+{x_{{3}}}^{2}
-{l}^{2}&-x_{{1}}x_{{0}}&-x_{{2}}x_{{0}}&-x_{{0}}x_{{3}}
\\  -x_{{1}}x_{{0}}&{x_{{0}}}^{2}-{x_{{2}}}^{2}-{x_{{
3}}}^{2}+{l}^{2}&x_{{2}}x_{{1}}&x_{{1}}x_{{3}}\\  -x_
{{2}}x_{{0}}&x_{{2}}x_{{1}}&{x_{{0}}}^{2}-{x_{{1}}}^{2}-{x_{{3}}}^{2}+
{l}^{2}&x_{{3}}x_{{2}}\\  -x_{{0}}x_{{3}}&x_{{1}}x_{{
3}}&x_{{3}}x_{{2}}&{x_{{0}}}^{2}-{x_{{1}}}^{2}-{x_{{2}}}^{2}+{l}^{2}
\end {array} \right] 
$$
where
$$d={x_{{0}}}^{2}-{x_{{1}}}^{2}-{x_{{2}}}^{2}-{x_{{3}}}^{2}+{l}^{2}$$
the metric is singular if $~d=0~$
