The shape of the catenary is the result of the requirement that internal forces are only acting tangential to the curve shape, and there is no shear or bending moment along the wire. If $H$ is the horizontal tension, and $V$ the vertical (shear) tension then you have $$ \frac{V}{H} =\frac{{\rm d}y}{{\rm d}x} $$
The segment's total weight is balanced by the difference in vertical forces (and hence zero internal shear forces).
$$ {\rm d}V = \lambda \sqrt{1+ \left(\tfrac{{\rm d}y}{{\rm d}x}\right)^2} {\rm d}x $$
Here is a post of mine about the development of the equations.
So now what can happen if you design a cable with built-in curvature to match the shape? well, essentially a thin wire has zero resistance to bending (an assumption for the catenary shape development since internal moments are zero). So the initial shape of the cable does not matter. In fact, it comes in a spool that is much more curved than when installed.
Or maybe you are thinking of replacing the stranded wire with a solid one of the correct shape, which has other problems. For one, the internal tensions are still going to be tangential only, but due to the extra weight, natural frequencies will be much lower and hence more prone to galloping vibrations. And the friction between the strands provides a level of vibration damping that protects the cable.
Also, the current capacity will be diminished as electrons like to travel on the surface of the metal and the stranded shape has a much higher surface to volume ratio.
Finally, mechanically speaking it would be harder for a solid wire to respond to changing environmental conditions (like temperature, ice and wind) which change the loading and the force balance shape.