Question

Identify the surface of the z² = 4 + 4r² by converting them into equations in the Cartesian form. Show the complete solutions.

Answer 1

$$z^2=4+4r^2=>z^2=4+4x^2+4y^2$$

$$-\dfrac{x^2}{1^2}-\dfrac{y^2}{1^2}+\dfrac{z^2}{2^2}=1$$

The  hyperboloid of two sheets.