Réitigh {l}{x^2/4+y^2/2=1}{x=my+1}

$$\left\{ \begin{array} { l } { \frac { x ^ { 2 } } { 4 } + \frac { y ^ { 2 } } { 2 } = 1 } \\ { x = m y + 1 } \end{array} \right.$$

Réitigh do x,y.

$$x=\frac{\sqrt{2}\left(-m\sqrt{2m^{2}+3}+\sqrt{2}\right)}{m^{2}+2}\text{, }y=-\frac{\sqrt{2\left(2m^{2}+3\right)}+m}{m^{2}+2}$$

$$x=\frac{\sqrt{2}\left(m\sqrt{2m^{2}+3}+\sqrt{2}\right)}{m^{2}+2}\text{, }y=\frac{\sqrt{2\left(2m^{2}+3\right)}-m}{m^{2}+2}$$

Céimeanna ag Baint Úsáid as Ionadú agus an bhFoirme Chearnach

Cuir an chéad cothromóid san áireamh. Iolraigh an dá thaobh den chothromóid faoi 4, an comhiolraí is lú de 4,2.

$$x^{2}+2y^{2}=4$$

Cuir an dara cothromóid san áireamh. Bain my ón dá thaobh.

$$x-my=1$$

Chun péire cothromóidí a réiteach ag baint úsáid as ionadú, réitigh ceann de na cothromóidí ar dtús le ceann de na hathróga a fháil. Ansin ionadaigh an toradh don athróg sin sa chothromóid eile.

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

Réitigh x+\left(-m\right)y=1 do x trí x ar an taobh clé den chomhartha ‘Cothrom le’ a aonrú.

$$x+\left(-m\right)y=1$$

Bain \left(-m\right)y ón dá thaobh den chothromóid.

$$x=my+1$$

Cuir x in aonad my+1 sa chothromóid eile, 2y^{2}+x^{2}=4.

$$2y^{2}+\left(my+1\right)^{2}=4$$

Cearnóg my+1.

$$2y^{2}+m^{2}y^{2}+2my+1=4$$

Suimigh 2y^{2} le m^{2}y^{2}?

$$\left(m^{2}+2\right)y^{2}+2my+1=4$$

Bain 4 ón dá thaobh den chothromóid.

$$\left(m^{2}+2\right)y^{2}+2my-3=0$$

Tá an chothromóid seo i bhfoirm chaighdeánach: ax^{2}+bx+c=0. Cuir 2+1m^{2} in ionad a, 1\times 1\times 2m in ionad b, agus -3 in ionad c san fhoirmle chearnach, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-2m±\sqrt{\left(2m\right)^{2}-4\left(m^{2}+2\right)\left(-3\right)}}{2\left(m^{2}+2\right)}$$

Cearnóg 1\times 1\times 2m.

$$y=\frac{-2m±\sqrt{4m^{2}-4\left(m^{2}+2\right)\left(-3\right)}}{2\left(m^{2}+2\right)}$$

Méadaigh -4 faoi 2+1m^{2}.

$$y=\frac{-2m±\sqrt{4m^{2}+\left(-4m^{2}-8\right)\left(-3\right)}}{2\left(m^{2}+2\right)}$$

Méadaigh -8-4m^{2} faoi -3.

$$y=\frac{-2m±\sqrt{4m^{2}+12m^{2}+24}}{2\left(m^{2}+2\right)}$$

Suimigh 4m^{2} le 24+12m^{2}?

$$y=\frac{-2m±\sqrt{16m^{2}+24}}{2\left(m^{2}+2\right)}$$

Tóg fréamh chearnach 24+16m^{2}.

$$y=\frac{-2m±2\sqrt{4m^{2}+6}}{2\left(m^{2}+2\right)}$$

Méadaigh 2 faoi 2+1m^{2}.

$$y=\frac{-2m±2\sqrt{4m^{2}+6}}{2m^{2}+4}$$

Réitigh an chothromóid y=\frac{-2m±2\sqrt{4m^{2}+6}}{2m^{2}+4} nuair is ionann ± agus plus. Suimigh -2m le 2\sqrt{6+4m^{2}}?

$$y=\frac{2\sqrt{4m^{2}+6}-2m}{2m^{2}+4}$$

Roinn -2m+2\sqrt{6+4m^{2}} faoi 4+2m^{2}.

$$y=\frac{\sqrt{4m^{2}+6}-m}{m^{2}+2}$$

Réitigh an chothromóid y=\frac{-2m±2\sqrt{4m^{2}+6}}{2m^{2}+4} nuair is ionann ± agus míneas. Dealaigh 2\sqrt{6+4m^{2}} ó -2m.

$$y=\frac{-2\sqrt{4m^{2}+6}-2m}{2m^{2}+4}$$

Roinn -2m-2\sqrt{6+4m^{2}} faoi 4+2m^{2}.

$$y=-\frac{\sqrt{4m^{2}+6}+m}{m^{2}+2}$$

Tá dhá réiteach ann do y: \frac{-m+\sqrt{6+4m^{2}}}{2+m^{2}} agus -\frac{m+\sqrt{6+4m^{2}}}{2+m^{2}}. Cuir y in aonad \frac{-m+\sqrt{6+4m^{2}}}{2+m^{2}} sa chothromóid eile x=my+1 chun an réiteach comhfhreagrach do x a shásaíonn an dá chothromóid a fháil.

$$x=m\times \frac{\sqrt{4m^{2}+6}-m}{m^{2}+2}+1$$

Méadaigh m faoi \frac{-m+\sqrt{6+4m^{2}}}{2+m^{2}}.

$$x=\frac{\sqrt{4m^{2}+6}-m}{m^{2}+2}m+1$$

Suimigh m\times \frac{-m+\sqrt{6+4m^{2}}}{2+m^{2}} le 1?

$$x=1+\frac{\sqrt{4m^{2}+6}-m}{m^{2}+2}m$$

Ansin cuir y in aonad -\frac{m+\sqrt{6+4m^{2}}}{2+m^{2}} sa chothromóid eile x=my+1 agus faigh réiteach chun an réiteach comhfhreagrach do x a shásaíonn an dá chothromóid a fháil.

$$x=m\left(-\frac{\sqrt{4m^{2}+6}+m}{m^{2}+2}\right)+1$$

Méadaigh m faoi -\frac{m+\sqrt{6+4m^{2}}}{2+m^{2}}.

$$x=\left(-\frac{\sqrt{4m^{2}+6}+m}{m^{2}+2}\right)m+1$$

Suimigh m\left(-\frac{m+\sqrt{6+4m^{2}}}{2+m^{2}}\right) le 1?

$$x=1+\left(-\frac{\sqrt{4m^{2}+6}+m}{m^{2}+2}\right)m$$

Tá an córas réitithe anois.

$$x=1+\frac{\sqrt{4m^{2}+6}-m}{m^{2}+2}m,y=\frac{\sqrt{4m^{2}+6}-m}{m^{2}+2}\text{ or }x=1+\left(-\frac{\sqrt{4m^{2}+6}+m}{m^{2}+2}\right)m,y=-\frac{\sqrt{4m^{2}+6}+m}{m^{2}+2}$$

Réitigh do x,y. (complex solution)

$$\left\{\begin{matrix}x=\frac{\sqrt{2}\left(-m\sqrt{2m^{2}+3}+\sqrt{2}\right)}{m^{2}+2}\text{, }y=-\frac{\sqrt{2\left(2m^{2}+3\right)}+m}{m^{2}+2}\text{; }x=\frac{\sqrt{2}\left(m\sqrt{2m^{2}+3}+\sqrt{2}\right)}{m^{2}+2}\text{, }y=\frac{\sqrt{2\left(2m^{2}+3\right)}-m}{m^{2}+2}\text{, }&m\neq -\sqrt{2}i\text{ and }m\neq \sqrt{2}i\\x=\frac{5}{2}=2.5\text{, }y=\frac{3}{2m}\text{, }&m=-\sqrt{2}i\text{ or }m=\sqrt{2}i\end{matrix}\right.$$