求解 x^2/-2+y^2/b^2=1

$$\frac { x ^ { 2 } } { - 2 } + \frac { y ^ { 2 } } { b ^ { 2 } } = 1$$

求解 b 的值 (复数求解)

$$\left\{\begin{matrix}b=-\sqrt{2}i\left(-x^{2}-2\right)^{-\frac{1}{2}}y\text{; }b=\sqrt{2}i\left(-x^{2}-2\right)^{-\frac{1}{2}}y\text{, }&y\neq 0\text{ and }x\neq -\sqrt{2}i\text{ and }x\neq \sqrt{2}i\\b\neq 0\text{, }&\left(x=\sqrt{2}i\text{ or }x=-\sqrt{2}i\right)\text{ and }y=0\end{matrix}\right.$$

求解 x 的值 (复数求解)

$$x=-\sqrt{2\times \left(\frac{y}{b}\right)^{2}-2}$$

$$x=\sqrt{2\times \left(\frac{y}{b}\right)^{2}-2}\text{, }b\neq 0$$

求解 b 的值

$$b=\sqrt{\frac{2}{x^{2}+2}}y$$

$$b=-\sqrt{\frac{2}{x^{2}+2}}y\text{, }y\neq 0$$

求解 x 的值

$$x=\frac{\sqrt{2\left(y^{2}-b^{2}\right)}}{|b|}$$

$$x=-\frac{\sqrt{2\left(y^{2}-b^{2}\right)}}{|b|}\text{, }b\neq 0\text{ and }|y|\geq |b|$$