$\dfrac{x^2}{4} + \dfrac{y^2}{4} + \dfrac{z^2}{9} = 1$

$\dfrac{z}{3} = \dfrac{x^2}{4} + \dfrac{y^2}{4}$

$\dfrac{z^2}{9} = \dfrac{x^2}{4} + \dfrac{y^2}{4}$

$-\dfrac{x^2}{4} - \dfrac{y^2}{4} + \dfrac{z^2}{9} = 1$

$\dfrac{z}{3} = \dfrac{x^2}{4} - \dfrac{y^2}{4}$