Réitigh {l}{y=kx+b}{x^2/4+y^2=1}

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

Réitigh do x,y.

$$x=-\frac{2\left(2bk+\sqrt{1+4k^{2}-b^{2}}\right)}{4k^{2}+1}\text{, }y=\frac{-2k\sqrt{1+4k^{2}-b^{2}}+b}{4k^{2}+1}$$

$$x=\frac{2\left(-2bk+\sqrt{1+4k^{2}-b^{2}}\right)}{4k^{2}+1}\text{, }y=\frac{2k\sqrt{1+4k^{2}-b^{2}}+b}{4k^{2}+1}\text{, }|k|\geq \frac{\sqrt{b^{2}-1}}{2}\text{ or }|b|<1$$

Céimeanna ag Baint Úsáid as Ionadú agus an bhFoirme Chearnach

Cuir an chéad cothromóid san áireamh. Bain kx ón dá thaobh.

$$y-kx=b$$

Cuir an dara cothromóid san áireamh. Méadaigh an dá thaobh den chothromóid faoi 4.

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

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.

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

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

$$y+\left(-k\right)x=b$$

Bain \left(-k\right)x ón dá thaobh den chothromóid.

$$y=kx+b$$

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

$$x^{2}+4\left(kx+b\right)^{2}=4$$

Cearnóg kx+b.

$$x^{2}+4\left(k^{2}x^{2}+2bkx+b^{2}\right)=4$$

Méadaigh 4 faoi k^{2}x^{2}+2bkx+b^{2}.

$$x^{2}+4k^{2}x^{2}+8bkx+4b^{2}=4$$

Suimigh x^{2} le 4k^{2}x^{2}?

$$\left(4k^{2}+1\right)x^{2}+8bkx+4b^{2}=4$$

Bain 4 ón dá thaobh den chothromóid.

$$\left(4k^{2}+1\right)x^{2}+8bkx+4b^{2}-4=0$$

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

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

Cearnóg 4\times 2kb.

$$x=\frac{-8bk±\sqrt{64b^{2}k^{2}-4\left(4k^{2}+1\right)\left(4b^{2}-4\right)}}{2\left(4k^{2}+1\right)}$$

Méadaigh -4 faoi 1+4k^{2}.

$$x=\frac{-8bk±\sqrt{64b^{2}k^{2}+\left(-16k^{2}-4\right)\left(4b^{2}-4\right)}}{2\left(4k^{2}+1\right)}$$

Méadaigh -4-16k^{2} faoi -4+4b^{2}.

$$x=\frac{-8bk±\sqrt{64b^{2}k^{2}-16\left(b^{2}-1\right)\left(4k^{2}+1\right)}}{2\left(4k^{2}+1\right)}$$

Suimigh 64k^{2}b^{2} le -16\left(1+4k^{2}\right)\left(b^{2}-1\right)?

$$x=\frac{-8bk±\sqrt{16+64k^{2}-16b^{2}}}{2\left(4k^{2}+1\right)}$$

Tóg fréamh chearnach -16b^{2}+64k^{2}+16.

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

Méadaigh 2 faoi 1+4k^{2}.

$$x=\frac{-8bk±4\sqrt{1+4k^{2}-b^{2}}}{8k^{2}+2}$$

Réitigh an chothromóid x=\frac{-8bk±4\sqrt{1+4k^{2}-b^{2}}}{8k^{2}+2} nuair is ionann ± agus plus. Suimigh -8kb le 4\sqrt{-b^{2}+4k^{2}+1}?

$$x=\frac{-8bk+4\sqrt{1+4k^{2}-b^{2}}}{8k^{2}+2}$$

Roinn -8bk+4\sqrt{-b^{2}+4k^{2}+1} faoi 2+8k^{2}.

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

Réitigh an chothromóid x=\frac{-8bk±4\sqrt{1+4k^{2}-b^{2}}}{8k^{2}+2} nuair is ionann ± agus míneas. Dealaigh 4\sqrt{-b^{2}+4k^{2}+1} ó -8kb.

$$x=\frac{-8bk-4\sqrt{1+4k^{2}-b^{2}}}{8k^{2}+2}$$

Roinn -8kb-4\sqrt{-b^{2}+4k^{2}+1} faoi 2+8k^{2}.

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

Tá dhá réiteach ann do x: \frac{2\left(-2bk+\sqrt{-b^{2}+4k^{2}+1}\right)}{1+4k^{2}} agus -\frac{2\left(2bk+\sqrt{-b^{2}+4k^{2}+1}\right)}{1+4k^{2}}. Cuir x in aonad \frac{2\left(-2bk+\sqrt{-b^{2}+4k^{2}+1}\right)}{1+4k^{2}} sa chothromóid eile y=kx+b chun an réiteach comhfhreagrach do y a shásaíonn an dá chothromóid a fháil.

$$y=k\times \frac{2\left(-2bk+\sqrt{1+4k^{2}-b^{2}}\right)}{4k^{2}+1}+b$$

Méadaigh k faoi \frac{2\left(-2bk+\sqrt{-b^{2}+4k^{2}+1}\right)}{1+4k^{2}}.

$$y=\frac{2\left(-2bk+\sqrt{1+4k^{2}-b^{2}}\right)}{4k^{2}+1}k+b$$

Ansin cuir x in aonad -\frac{2\left(2bk+\sqrt{-b^{2}+4k^{2}+1}\right)}{1+4k^{2}} sa chothromóid eile y=kx+b agus faigh réiteach chun an réiteach comhfhreagrach do y a shásaíonn an dá chothromóid a fháil.

$$y=k\left(-\frac{2\left(2bk+\sqrt{1+4k^{2}-b^{2}}\right)}{4k^{2}+1}\right)+b$$

Méadaigh k faoi -\frac{2\left(2bk+\sqrt{-b^{2}+4k^{2}+1}\right)}{1+4k^{2}}.

$$y=\left(-\frac{2\left(2bk+\sqrt{1+4k^{2}-b^{2}}\right)}{4k^{2}+1}\right)k+b$$

Tá an córas réitithe anois.

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

Réitigh do x,y. (complex solution)

$$\left\{\begin{matrix}x=-\frac{2\left(2bk+\sqrt{1+4k^{2}-b^{2}}\right)}{4k^{2}+1}\text{, }y=\frac{-2k\sqrt{1+4k^{2}-b^{2}}+b}{4k^{2}+1}\text{; }x=\frac{2\left(-2bk+\sqrt{1+4k^{2}-b^{2}}\right)}{4k^{2}+1}\text{, }y=\frac{2k\sqrt{1+4k^{2}-b^{2}}+b}{4k^{2}+1}\text{, }&k\neq -\frac{1}{2}i\text{ and }k\neq \frac{1}{2}i\\x=-\frac{b^{2}-1}{2bk}\text{, }y=\frac{b^{2}+1}{2b}\text{, }&b\neq 0\text{ and }\left(k=-\frac{1}{2}i\text{ or }k=\frac{1}{2}i\right)\end{matrix}\right.$$