{l}{x^2+2y^2=8}{y=m(x-2)+sqrt{2}} ਨੂੰ ਹੱਲ ਕਰੋ

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

x, y ਲਈ ਹਲ ਕਰੋ

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

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

ਸਬਸੀਟਿਉਸ਼ਨ ਅਤੇ ਵਰਗਾਕਾਰ ਸੂਤਰ ਨੂੰ ਵਰਤ ਰਹੇ ਸਟੈਪਸ

ਦੂਜੇ ਸਮੀਕਰਨ 'ਤੇ ਵਿਚਾਰ ਕਰੋ। m ਨੂੰ x-2 ਨਾਲ ਗੁਣਾ ਕਰਨ ਲਈ ਡਿਸਟ੍ਰੀਬਿਉਟਿਵ ਪ੍ਰੋਪਰਟੀ ਨੂੰ ਵਰਤੋਂ।

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

ਦੂਜੇ ਸਮੀਕਰਨ x^{2}+2y^{2}=8 ਵਿੱਚ, y ਲਈ mx-2m+\sqrt{2} ਨੂੰ ਬਦਲ ਦਿਓ।

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

mx-2m+\sqrt{2} ਦਾ ਵਰਗ ਕਰੋ।

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

2 ਨੂੰ m^{2}x^{2}+2m\left(-2m+\sqrt{2}\right)x+\left(-2m+\sqrt{2}\right)^{2} ਵਾਰ ਗੁਣਾ ਕਰੋ।

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

x^{2} ਨੂੰ 2m^{2}x^{2} ਵਿੱਚ ਜੋੜੋ।

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

ਸਮੀਕਰਨ ਦੇ ਦੋਹਾਂ ਪਾਸਿਆਂ ਤੋਂ 8 ਨੂੰ ਘਟਾਓ।

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

ਇਹ ਸਮੀਕਰਨ ਮਿਆਰੀ ਰੂਪ ਵਿੱਚ ਹੈ: ax^{2}+bx+c=0. ਵਰਗਾਤਮਕ ਸੂਤਰ \frac{-b±\sqrt{b^{2}-4ac}}{2a} ਵਿੱਚ 1+2m^{2} ਨੂੰ a ਲਈ, 2\times 2m\left(-2m+\sqrt{2}\right) ਨੂੰ b ਲਈ, ਅਤੇ -4+8m^{2}-8m\sqrt{2} ਨੂੰ c ਲਈ ਬਦਲ ਦਿਓ।

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

2\times 2m\left(-2m+\sqrt{2}\right) ਦਾ ਵਰਗ ਕਰੋ।

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

-4 ਨੂੰ 1+2m^{2} ਵਾਰ ਗੁਣਾ ਕਰੋ।

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

-4-8m^{2} ਨੂੰ -4+8m^{2}-8m\sqrt{2} ਵਾਰ ਗੁਣਾ ਕਰੋ।

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

16m^{2}\left(-2m+\sqrt{2}\right)^{2} ਨੂੰ 16+32m\sqrt{2}-64m^{4}+64m^{3}\sqrt{2} ਵਿੱਚ ਜੋੜੋ।

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

16+32m^{2}+32m\sqrt{2} ਦਾ ਵਰਗ ਮੂਲ ਲਓ।

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

2 ਨੂੰ 1+2m^{2} ਵਾਰ ਗੁਣਾ ਕਰੋ।

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

ਹੁਣ, ਸਮੀਕਰਨ x=\frac{-4m\left(-2m+\sqrt{2}\right)±4\sqrt{2m^{2}+2\sqrt{2}m+1}}{4m^{2}+2} ਨੂੰ ਸੁਲਝਾਓ ਜਦੋਂ ± ਪਲੱਸ ਹੁੰਦਾ ਹੈ। -4m\left(-2m+\sqrt{2}\right) ਨੂੰ 4\sqrt{1+2m^{2}+2m\sqrt{2}} ਵਿੱਚ ਜੋੜੋ।

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

-4m\left(-2m+\sqrt{2}\right)+4\sqrt{1+2m^{2}+2m\sqrt{2}} ਨੂੰ 2+4m^{2} ਦੇ ਨਾਲ ਤਕਸੀਮ ਕਰੋ।

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

ਹੁਣ, ਸਮੀਕਰਨ x=\frac{-4m\left(-2m+\sqrt{2}\right)±4\sqrt{2m^{2}+2\sqrt{2}m+1}}{4m^{2}+2} ਨੂੰ ਸੁਲਝਾਓ ਜਦੋਂ ± ਮਾਈਨਸ ਹੁੰਦਾ ਹੈ। -4m\left(-2m+\sqrt{2}\right) ਵਿੱਚੋਂ 4\sqrt{1+2m^{2}+2m\sqrt{2}} ਨੂੰ ਘਟਾਓ।

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

8m^{2}-4m\sqrt{2}-4\sqrt{1+2m^{2}+2m\sqrt{2}} ਨੂੰ 2+4m^{2} ਦੇ ਨਾਲ ਤਕਸੀਮ ਕਰੋ।

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

x ਲਈ ਦੋ ਹਲ: \frac{2\left(2m^{2}-m\sqrt{2}+\sqrt{2m^{2}+1+2m\sqrt{2}}\right)}{1+2m^{2}} ਅਤੇ \frac{2\left(2m^{2}-m\sqrt{2}-\sqrt{2m^{2}+1+2m\sqrt{2}}\right)}{1+2m^{2}} ਹੁੰਦੇ ਹਨ। ਸਮੀਕਰਨ y=mx-2m+\sqrt{2} ਵਿੱਚ x ਲਈ \frac{2\left(2m^{2}-m\sqrt{2}+\sqrt{2m^{2}+1+2m\sqrt{2}}\right)}{1+2m^{2}} ਨੂੰ ਬਦਲੋ ਅਤੇ y ਲਈ ਸੰਗਤ ਹਲ ਕੱਢੋ ਜੋ ਦੋਵੇਂ ਸਮੀਕਰਨਾਂ ਨੂੰ ਸੰਤੁਸ਼ਟ ਕਰੇ।

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

m ਨੂੰ \frac{2\left(2m^{2}-m\sqrt{2}+\sqrt{2m^{2}+1+2m\sqrt{2}}\right)}{1+2m^{2}} ਵਾਰ ਗੁਣਾ ਕਰੋ।

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

ਹੁਣ, ਸਮੀਕਰਨ y=mx-2m+\sqrt{2} ਵਿੱਚ x ਲਈ \frac{2\left(2m^{2}-m\sqrt{2}-\sqrt{2m^{2}+1+2m\sqrt{2}}\right)}{1+2m^{2}} ਨੂੰ ਬਦਲੋ ਅਤੇ y ਲਈ ਸੰਗਤ ਹਲ ਕੱਢਣ ਲਈ ਹਲ ਕਰੋ, ਜੋ ਦੋਵੇਂ ਸਮੀਕਰਨਾਂ ਨੂੰ ਸੰਤੁਸ਼ਟ ਕਰੇ।

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

m ਨੂੰ \frac{2\left(2m^{2}-m\sqrt{2}-\sqrt{2m^{2}+1+2m\sqrt{2}}\right)}{1+2m^{2}} ਵਾਰ ਗੁਣਾ ਕਰੋ।

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

ਸਿਸਟਮ ਹੁਣ ਸੁਲਝ ਗਿਆ ਹੈ।

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

x, y ਲਈ ਹਲ ਕਰੋ (ਜਟਿਲ ਹੱਲ)

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