समाधान {l}{y=x-p/2}{y^2=2p^2x}

$$\left\{ \begin{array} { l } { y = x - \frac { p } { 2 } } \\ { y ^ { 2 } = 2 p ^ { 2 } x } \end{array} \right.$$

x, y को लागि हल गर्नुहोस् (complex solution)

$$x=-\sqrt{p^{4}+p^{3}}+p^{2}+\frac{p}{2}\text{, }y=-\sqrt{p^{4}+p^{3}}+p^{2}$$

$$x=\sqrt{p^{4}+p^{3}}+p^{2}+\frac{p}{2}\text{, }y=\sqrt{p^{4}+p^{3}}+p^{2}$$

x, y को लागि हल गर्नुहोस्

$$\left\{\begin{matrix}x=-p\sqrt{p^{2}+p}+p^{2}+\frac{p}{2}\text{, }y=-\sqrt{p+1}p^{\frac{3}{2}}+p^{2}\text{; }x=p\sqrt{p^{2}+p}+p^{2}+\frac{p}{2}\text{, }y=\sqrt{p+1}p^{\frac{3}{2}}+p^{2}\text{, }&p\geq 0\\x=-\sqrt{p^{4}+p^{3}}+p^{2}+\frac{p}{2}\text{, }y=-\sqrt{p^{4}+p^{3}}+p^{2}\text{; }x=\sqrt{p^{4}+p^{3}}+p^{2}+\frac{p}{2}\text{, }y=\sqrt{p^{4}+p^{3}}+p^{2}\text{, }&p\leq -1\end{matrix}\right.$$