Given, $$\varphi = a(x^{2} + y^{2}) + bz^{2}$$
So, $$\vec {E} = -\vec {\bigtriangledown}\varphi = -[2ax \vec {i} + 2ay\ vec {j} + 2bz \vec {k}]$$
Hence $$|\vec {E}| = 2\sqrt {a^{2}(x^{2} + y^{2}) + b^{2}z^{2}}$$
Shape of the equipotential surface:
Put $$\vec {\rho} = x\vec {i} + y\vec {j}$$ or $$\rho^{2} = x^{2} + y^{2}$$
Then the equipotential surface has the equation
$$a\rho^{2} + bz^{2} = constant = \varphi$$
If $$a > 0, b > 0$$ then $$\varphi > 0$$ and the equation of the equipotential surface is
$$\vec {\rho}{\varphi/a} + \dfrac {z^{2}}{\varphi/b} = 1$$
which is an ellipse in $$\rho, z$$ coordinates. In three dimensions the surface is an ellipsoid of revolution with semi-axis $$\sqrt {\varphi/a}, \sqrt {\varphi/a}, \sqrt {\varphi/b}$$.
If $$a > 0, b < 0$$ then $$\varphi$$ can be $$\geq 0$$. If $$\varphi > 0$$ then the equation is
$$\dfrac {\rho^{2}}{\varphi/a} - \dfrac {z^{2}}{\varphi/|b|} = 1$$
This is a single cavity hyperboloiod of revolution about $$z$$ axis. If $$\varphi = 0$$ then
$$a\rho^{2} - |b|z^{2} = 0$$ or $$z = \pm \sqrt {\dfrac {a}{|b|}} \rho$$
is the equation of a right circular cone.
If $$\varphi < 0$$ then the equation can be written as
$$|b|z^{2} - a\rho^{2} = |\varphi|$$
or $$\dfrac {z^{2}}{|\varphi| / |b|} - \dfrac {\rho^{2}}{|\varphi|/a} = 1$$
This is a two cavity hyperboloid of revolution about z-axis.