{l}{y^2=2px}{x^2+y^2=16} ಪರಿಹರಿಸಿ

$$\left. \begin{array} { l } { y ^ { 2 } = 2 p x } \\ { x ^ { 2 } + y ^ { 2 } = 16 } \end{array} \right.$$

x, y ಪರಿಹರಿಸಿ (ಸಂಕೀರ್ಣ ಪರಿಹಾರ)

$$x=-\left(\sqrt{p^{2}+16}+p\right)\text{, }y=i\sqrt{p}\sqrt{2\left(\sqrt{p^{2}+16}+p\right)}$$

$$x=-\left(\sqrt{p^{2}+16}+p\right)\text{, }y=-i\sqrt{p}\sqrt{2\left(\sqrt{p^{2}+16}+p\right)}$$

$$x=\sqrt{p^{2}+16}-p\text{, }y=\sqrt{p}\sqrt{2\left(\sqrt{p^{2}+16}-p\right)}$$

$$x=\sqrt{p^{2}+16}-p\text{, }y=-\sqrt{p}\sqrt{2\left(\sqrt{p^{2}+16}-p\right)}$$

x, y ಪರಿಹರಿಸಿ

$$\left\{\begin{matrix}x=-\left(\sqrt{p^{2}+16}+p\right)\text{, }y=-\sqrt{-2p\left(\sqrt{p^{2}+16}+p\right)}\text{; }x=-\left(\sqrt{p^{2}+16}+p\right)\text{, }y=\sqrt{-2p\left(\sqrt{p^{2}+16}+p\right)}\text{, }&p\leq 0\\x=\sqrt{p^{2}+16}-p\text{, }y=-\sqrt{2p\left(\sqrt{p^{2}+16}-p\right)}\text{; }x=\sqrt{p^{2}+16}-p\text{, }y=\sqrt{2p\left(\sqrt{p^{2}+16}-p\right)}\text{, }&p\geq 0\end{matrix}\right.$$