y=mx+a/m పరిష్కరించండి

$$y = m x + \frac { a } { m }$$

mని పరిష్కరించండి (సంకీర్ణ పరిష్కారం)

$$\left\{\begin{matrix}m=\frac{\sqrt{y^{2}-4ax}+y}{2x}\text{, }&\left(a\neq 0\text{ and }x\neq 0\right)\text{ or }\left(arg(y)<\pi \text{ and }y\neq 0\text{ and }x\neq 0\right)\\m=-\frac{\sqrt{y^{2}-4ax}-y}{2x}\text{, }&\left(a\neq 0\text{ and }x\neq 0\right)\text{ or }\left(|\frac{arg(y^{2})}{2}-arg(y)|\geq \pi \text{ and }y\neq 0\text{ and }x\neq 0\right)\\m=\frac{a}{y}\text{, }&a\neq 0\text{ and }y\neq 0\text{ and }x=0\\m\neq 0\text{, }&x=0\text{ and }y=0\text{ and }a=0\end{matrix}\right.$$

aని పరిష్కరించండి

$$a=-m\left(mx-y\right)$$

$$m\neq 0$$

mని పరిష్కరించండి

$$\left\{\begin{matrix}m=\frac{\sqrt{y^{2}-4ax}+y}{2x}\text{, }&\left(a>0\text{ and }x<0\right)\text{ or }\left(x>0\text{ and }a<0\right)\text{ or }\left(y\geq 2\sqrt{ax}\text{ and }y>0\text{ and }x<0\text{ and }a\leq 0\right)\text{ or }\left(y\geq 2\sqrt{ax}\text{ and }y>0\text{ and }x>0\text{ and }a\geq 0\right)\text{ or }\left(x\neq 0\text{ and }a\neq 0\text{ and }|y|\geq 2\sqrt{ax}\right)\\m=-\frac{\sqrt{y^{2}-4ax}-y}{2x}\text{, }&\left(x>0\text{ and }a<0\right)\text{ or }\left(a>0\text{ and }x<0\right)\text{ or }\left(y\leq -2\sqrt{ax}\text{ and }y<0\text{ and }x<0\text{ and }a\leq 0\right)\text{ or }\left(y\leq -2\sqrt{ax}\text{ and }y<0\text{ and }x>0\text{ and }a\geq 0\right)\text{ or }\left(x\neq 0\text{ and }a\neq 0\text{ and }|y|\geq 2\sqrt{ax}\right)\\m=\frac{a}{y}\text{, }&a\neq 0\text{ and }y\neq 0\text{ and }x=0\\m\neq 0\text{, }&x=0\text{ and }y=0\text{ and }a=0\end{matrix}\right.$$