{l}{x^2+2y^2=8}{y=m(x-2)+sqrt{2}} ಪರಿಹರಿಸಿ

$$\left. \begin{array} { l } { x ^ { 2 } + 2 y ^ { 2 } = 8 } \\ { y = m ( x - 2 ) + \sqrt { 2 } } \end{array} \right.$$

x, y ಪರಿಹರಿಸಿ

$$x=\frac{2\left(2m^{2}-\sqrt{2m^{2}+2\sqrt{2}m+1}-\sqrt{2}m\right)}{2m^{2}+1}\text{, }y=\frac{\sqrt{2}\left(-2m|\frac{\sqrt{2}\left(\sqrt{2}m+1\right)}{2}|-\sqrt{2}m+1\right)}{2m^{2}+1}$$

$$x=\frac{2\left(2m^{2}+\sqrt{2m^{2}+2\sqrt{2}m+1}-\sqrt{2}m\right)}{2m^{2}+1}\text{, }y=\frac{\sqrt{2}\left(2m|\frac{\sqrt{2}\left(\sqrt{2}m+1\right)}{2}|-\sqrt{2}m+1\right)}{2m^{2}+1}$$

ಪರ್ಯಾಯ ಮತ್ತು ವರ್ಗೀಯ ಸೂತ್ರ ಬಳಸುವಿಕೆ ಕ್ರಮಗಳು

ಎರಡನೆಯ ಸಮೀಕರಣ ಪರಿಗಣಿಸಿ. x-2 ದಿಂದ m ಗುಣಿಸಲು ವಿಭಾಜಕ ಗುಣವನ್ನು ಬಳಸಿ.

$$y=mx-2m+\sqrt{2}$$

ಇತರ ಸಮೀಕರಣ x^{2}+2y^{2}=8 ನಲ್ಲಿ y ಗಾಗಿ mx-2m+\sqrt{2} ಬದಲಿಸಿ.

$$x^{2}+2\left(mx-2m+\sqrt{2}\right)^{2}=8$$

ವರ್ಗ mx-2m+\sqrt{2}.

$$x^{2}+2\left(m^{2}x^{2}+2m\left(-2m+\sqrt{2}\right)x+\left(-2m+\sqrt{2}\right)^{2}\right)=8$$

m^{2}x^{2}+2m\left(-2m+\sqrt{2}\right)x+\left(-2m+\sqrt{2}\right)^{2} ಅನ್ನು 2 ಬಾರಿ ಗುಣಿಸಿ.

$$x^{2}+2m^{2}x^{2}+4m\left(-2m+\sqrt{2}\right)x+2\left(-2m+\sqrt{2}\right)^{2}=8$$

2m^{2}x^{2} ಗೆ x^{2} ಸೇರಿಸಿ.

$$\left(2m^{2}+1\right)x^{2}+4m\left(-2m+\sqrt{2}\right)x+2\left(-2m+\sqrt{2}\right)^{2}=8$$

ಸಮೀಕರಣದ ಎರಡೂ ಕಡೆಗಳಿಂದ 8 ಕಳೆಯಿರಿ.

$$\left(2m^{2}+1\right)x^{2}+4m\left(-2m+\sqrt{2}\right)x+2\left(-2m+\sqrt{2}\right)^{2}-8=0$$

ಈ ಸಮೀಕರಣವು ಪ್ರಮಾಣಿತ ಫಾರ್ಮ್‌ನಲ್ಲಿದೆ: ax^{2}+bx+c=0. ವರ್ಗ ಸೂತ್ರ \frac{-b±\sqrt{b^{2}-4ac}}{2a} ನಲ್ಲಿ a ಗೆ 1+2m^{2}, b ಗೆ 2\times 2m\left(-2m+\sqrt{2}\right) ಮತ್ತು c ಗೆ -4+8m^{2}-8m\sqrt{2} ಬದಲಿಸಿ.

$$x=\frac{-4m\left(-2m+\sqrt{2}\right)±\sqrt{\left(4m\left(-2m+\sqrt{2}\right)\right)^{2}-4\left(2m^{2}+1\right)\left(8m^{2}-8\sqrt{2}m-4\right)}}{2\left(2m^{2}+1\right)}$$

ವರ್ಗ 2\times 2m\left(-2m+\sqrt{2}\right).

$$x=\frac{-4m\left(-2m+\sqrt{2}\right)±\sqrt{16m^{2}\left(-2m+\sqrt{2}\right)^{2}-4\left(2m^{2}+1\right)\left(8m^{2}-8\sqrt{2}m-4\right)}}{2\left(2m^{2}+1\right)}$$

1+2m^{2} ಅನ್ನು -4 ಬಾರಿ ಗುಣಿಸಿ.

$$x=\frac{-4m\left(-2m+\sqrt{2}\right)±\sqrt{16m^{2}\left(-2m+\sqrt{2}\right)^{2}+\left(-8m^{2}-4\right)\left(8m^{2}-8\sqrt{2}m-4\right)}}{2\left(2m^{2}+1\right)}$$

-4+8m^{2}-8m\sqrt{2} ಅನ್ನು -4-8m^{2} ಬಾರಿ ಗುಣಿಸಿ.

$$x=\frac{-4m\left(-2m+\sqrt{2}\right)±\sqrt{16m^{2}\left(-2m+\sqrt{2}\right)^{2}-64m^{4}+64\sqrt{2}m^{3}+32\sqrt{2}m+16}}{2\left(2m^{2}+1\right)}$$

16+32m\sqrt{2}-64m^{4}+64m^{3}\sqrt{2} ಗೆ 16m^{2}\left(-2m+\sqrt{2}\right)^{2} ಸೇರಿಸಿ.

$$x=\frac{-4m\left(-2m+\sqrt{2}\right)±\sqrt{32m^{2}+32\sqrt{2}m+16}}{2\left(2m^{2}+1\right)}$$

16+32m^{2}+32m\sqrt{2} ನ ವರ್ಗಮೂಲವನ್ನು ತೆಗೆದುಕೊಳ್ಳಿ.

$$x=\frac{-4m\left(-2m+\sqrt{2}\right)±4\sqrt{2m^{2}+2\sqrt{2}m+1}}{2\left(2m^{2}+1\right)}$$

1+2m^{2} ಅನ್ನು 2 ಬಾರಿ ಗುಣಿಸಿ.

$$x=\frac{-4m\left(-2m+\sqrt{2}\right)±4\sqrt{2m^{2}+2\sqrt{2}m+1}}{4m^{2}+2}$$

± ಎನ್ನುವುದು ಧನಾತ್ಮಕವಾಗಿರುವಾಗ x=\frac{-4m\left(-2m+\sqrt{2}\right)±4\sqrt{2m^{2}+2\sqrt{2}m+1}}{4m^{2}+2} ಸಮೀಕರಣವನ್ನು ಇದೀಗ ಪರಿಹರಿಸಿ. 4\sqrt{1+2m^{2}+2m\sqrt{2}} ಗೆ -4m\left(-2m+\sqrt{2}\right) ಸೇರಿಸಿ.

$$x=\frac{-4m\left(-2m+\sqrt{2}\right)+4\sqrt{2m^{2}+2\sqrt{2}m+1}}{4m^{2}+2}$$

2+4m^{2} ದಿಂದ -4m\left(-2m+\sqrt{2}\right)+4\sqrt{1+2m^{2}+2m\sqrt{2}} ಭಾಗಿಸಿ.

$$x=\frac{2\left(2m^{2}+\sqrt{2m^{2}+2\sqrt{2}m+1}-\sqrt{2}m\right)}{2m^{2}+1}$$

± ಎನ್ನುವುದು ಋಣಾತ್ಮಕವಾಗಿರುವಾಗ x=\frac{-4m\left(-2m+\sqrt{2}\right)±4\sqrt{2m^{2}+2\sqrt{2}m+1}}{4m^{2}+2} ಸಮೀಕರಣವನ್ನು ಇದೀಗ ಪರಿಹರಿಸಿ. -4m\left(-2m+\sqrt{2}\right) ದಿಂದ 4\sqrt{1+2m^{2}+2m\sqrt{2}} ಕಳೆಯಿರಿ.

$$x=\frac{8m^{2}-4\sqrt{2m^{2}+2\sqrt{2}m+1}-4\sqrt{2}m}{4m^{2}+2}$$

2+4m^{2} ದಿಂದ 8m^{2}-4m\sqrt{2}-4\sqrt{1+2m^{2}+2m\sqrt{2}} ಭಾಗಿಸಿ.

$$x=\frac{2\left(2m^{2}-\sqrt{2m^{2}+2\sqrt{2}m+1}-\sqrt{2}m\right)}{2m^{2}+1}$$

x ಗೆ ಎರಡು ಪರಿಹಾರಗಳಿವೆ: \frac{2\left(2m^{2}-m\sqrt{2}+\sqrt{2m^{2}+1+2m\sqrt{2}}\right)}{1+2m^{2}} ಮತ್ತು \frac{2\left(2m^{2}-m\sqrt{2}-\sqrt{2m^{2}+1+2m\sqrt{2}}\right)}{1+2m^{2}}. ಎರಡೂ ಸಮೀಕರಣಗಳನ್ನು ತೃಪ್ತಿಗೊಳಿಸುವ y ಗೆ ಅನುಗುಣವಾದ ಪರಿಹಾರ ಹುಡುಕಲು y=mx-2m+\sqrt{2} ಸಮೀಕರಣದಲ್ಲಿ x ಗಾಗಿ \frac{2\left(2m^{2}-m\sqrt{2}+\sqrt{2m^{2}+1+2m\sqrt{2}}\right)}{1+2m^{2}} ಬದಲಿಸಿ.

$$y=m\times \frac{2\left(2m^{2}+\sqrt{2m^{2}+2\sqrt{2}m+1}-\sqrt{2}m\right)}{2m^{2}+1}-2m+\sqrt{2}$$

\frac{2\left(2m^{2}-m\sqrt{2}+\sqrt{2m^{2}+1+2m\sqrt{2}}\right)}{1+2m^{2}} ಅನ್ನು m ಬಾರಿ ಗುಣಿಸಿ.

$$y=\frac{2\left(2m^{2}+\sqrt{2m^{2}+2\sqrt{2}m+1}-\sqrt{2}m\right)}{2m^{2}+1}m-2m+\sqrt{2}$$

ಇದೀಗ y=mx-2m+\sqrt{2} ಸಮೀಕರಣದಲ್ಲಿ x ಗಾಗಿ \frac{2\left(2m^{2}-m\sqrt{2}-\sqrt{2m^{2}+1+2m\sqrt{2}}\right)}{1+2m^{2}} ಅನ್ನು ಬದಲಿಸಿ ಹಾಗೂ ಎರಡೂ ಸಮೀಕರಣಗಳನ್ನು ಪೂರೈಸುವ y ಗೆ ಅನುಗುಣವಾದ ಪರಿಹಾರ ಹುಡುಕಲು ಪರಿಹರಿಸಿ.

$$y=m\times \frac{2\left(2m^{2}-\sqrt{2m^{2}+2\sqrt{2}m+1}-\sqrt{2}m\right)}{2m^{2}+1}-2m+\sqrt{2}$$

\frac{2\left(2m^{2}-m\sqrt{2}-\sqrt{2m^{2}+1+2m\sqrt{2}}\right)}{1+2m^{2}} ಅನ್ನು m ಬಾರಿ ಗುಣಿಸಿ.

$$y=\frac{2\left(2m^{2}-\sqrt{2m^{2}+2\sqrt{2}m+1}-\sqrt{2}m\right)}{2m^{2}+1}m-2m+\sqrt{2}$$

ಸಿಸ್ಟಂ ಅನ್ನು ಇದೀಗ ಪರಿಹರಿಸಲಾಗಿದೆ.

$$y=\frac{2\left(2m^{2}+\sqrt{2m^{2}+2\sqrt{2}m+1}-\sqrt{2}m\right)}{2m^{2}+1}m-2m+\sqrt{2},x=\frac{2\left(2m^{2}+\sqrt{2m^{2}+2\sqrt{2}m+1}-\sqrt{2}m\right)}{2m^{2}+1}\text{ or }y=\frac{2\left(2m^{2}-\sqrt{2m^{2}+2\sqrt{2}m+1}-\sqrt{2}m\right)}{2m^{2}+1}m-2m+\sqrt{2},x=\frac{2\left(2m^{2}-\sqrt{2m^{2}+2\sqrt{2}m+1}-\sqrt{2}m\right)}{2m^{2}+1}$$

x, y ಪರಿಹರಿಸಿ (ಸಂಕೀರ್ಣ ಪರಿಹಾರ)

$$\left\{\begin{matrix}x=\frac{2\left(2m^{2}-\sqrt{2m^{2}+2\sqrt{2}m+1}-\sqrt{2}m\right)}{2m^{2}+1}\text{, }y=\frac{\sqrt{2}\left(-m\sqrt{2\left(\sqrt{2}m+1\right)^{2}}-\sqrt{2}m+1\right)}{2m^{2}+1}\text{; }x=\frac{2\left(2m^{2}+\sqrt{2m^{2}+2\sqrt{2}m+1}-\sqrt{2}m\right)}{2m^{2}+1}\text{, }y=\frac{\sqrt{2}\left(m\sqrt{2\left(\sqrt{2}m+1\right)^{2}}-\sqrt{2}m+1\right)}{2m^{2}+1}\text{, }&m\neq -\frac{\sqrt{2}i}{2}\text{ and }m\neq \frac{\sqrt{2}i}{2}\\x=-\frac{\left(-2m+\sqrt{2}\right)^{2}-4}{2m\left(-2m+\sqrt{2}\right)}\text{, }y=\frac{2m^{2}-2\sqrt{2}m+3}{-2m+\sqrt{2}}\text{, }&m=-\frac{\sqrt{2}i}{2}\text{ or }m=\frac{\sqrt{2}i}{2}\end{matrix}\right.$$