Solve {l}{x^2/4+y^2/2=1}{y=k(x-1)}

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

Solve for x, y

$$x=\frac{\sqrt{2}\left(\sqrt{2}k^{2}-\sqrt{3k^{2}+2}\right)}{2k^{2}+1}\text{, }y=-\frac{k\left(\sqrt{2\left(3k^{2}+2\right)}+1\right)}{2k^{2}+1}$$

$$x=\frac{\sqrt{2}\left(\sqrt{2}k^{2}+\sqrt{3k^{2}+2}\right)}{2k^{2}+1}\text{, }y=\frac{k\left(\sqrt{2\left(3k^{2}+2\right)}-1\right)}{2k^{2}+1}$$

Steps Using Substitution and Quadratic Formula

Consider the first equation. Multiply both sides of the equation by 4, the least common multiple of 4,2.

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

Consider the second equation. Use the distributive property to multiply k by x-1.

$$y=kx-k$$

Substitute kx-k for y in the other equation, x^{2}+2y^{2}=4.

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

Square kx-k.

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

Multiply 2 times k^{2}x^{2}+2\left(-k\right)kx+\left(-k\right)^{2}.

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

Add x^{2} to 2k^{2}x^{2}.

$$\left(2k^{2}+1\right)x^{2}+4\left(-k\right)kx+2\left(-k\right)^{2}=4$$

Subtract 4 from both sides of the equation.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+2k^{2} for a, 2\times 2k\left(-k\right) for b, and 2k^{2}-4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square 2\times 2k\left(-k\right).

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

Multiply -4 times 1+2k^{2}.

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

Multiply -4-8k^{2} times 2k^{2}-4.

$$x=\frac{-4\left(-k\right)k±\sqrt{16k^{4}+16+24k^{2}-16k^{4}}}{2\left(2k^{2}+1\right)}$$

Add 16k^{4} to 16+24k^{2}-16k^{4}.

$$x=\frac{-4\left(-k\right)k±\sqrt{24k^{2}+16}}{2\left(2k^{2}+1\right)}$$

Take the square root of 24k^{2}+16.

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

Multiply 2 times 1+2k^{2}.

$$x=\frac{4k^{2}±2\sqrt{6k^{2}+4}}{4k^{2}+2}$$

Now solve the equation x=\frac{4k^{2}±2\sqrt{6k^{2}+4}}{4k^{2}+2} when ± is plus. Add 4k^{2} to 2\sqrt{6k^{2}+4}.

$$x=\frac{4k^{2}+2\sqrt{6k^{2}+4}}{4k^{2}+2}$$

Divide 4k^{2}+2\sqrt{6k^{2}+4} by 2+4k^{2}.

$$x=\frac{\sqrt{2}\left(\sqrt{2}k^{2}+\sqrt{3k^{2}+2}\right)}{2k^{2}+1}$$

Now solve the equation x=\frac{4k^{2}±2\sqrt{6k^{2}+4}}{4k^{2}+2} when ± is minus. Subtract 2\sqrt{6k^{2}+4} from 4k^{2}.

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

Divide 4k^{2}-2\sqrt{6k^{2}+4} by 2+4k^{2}.

$$x=\frac{\sqrt{2}\left(\sqrt{2}k^{2}-\sqrt{3k^{2}+2}\right)}{2k^{2}+1}$$

There are two solutions for x: \frac{\sqrt{2}\left(\sqrt{2}k^{2}+\sqrt{3k^{2}+2}\right)}{1+2k^{2}} and \frac{\sqrt{2}\left(\sqrt{2}k^{2}-\sqrt{3k^{2}+2}\right)}{1+2k^{2}}. Substitute \frac{\sqrt{2}\left(\sqrt{2}k^{2}+\sqrt{3k^{2}+2}\right)}{1+2k^{2}} for x in the equation y=kx-k to find the corresponding solution for y that satisfies both equations.

$$y=k\times \frac{\sqrt{2}\left(\sqrt{2}k^{2}+\sqrt{3k^{2}+2}\right)}{2k^{2}+1}-k$$

Multiply k times \frac{\sqrt{2}\left(\sqrt{2}k^{2}+\sqrt{3k^{2}+2}\right)}{1+2k^{2}}.

$$y=\frac{\sqrt{2}\left(\sqrt{2}k^{2}+\sqrt{3k^{2}+2}\right)}{2k^{2}+1}k-k$$

Now substitute \frac{\sqrt{2}\left(\sqrt{2}k^{2}-\sqrt{3k^{2}+2}\right)}{1+2k^{2}} for x in the equation y=kx-k and solve to find the corresponding solution for y that satisfies both equations.

$$y=k\times \frac{\sqrt{2}\left(\sqrt{2}k^{2}-\sqrt{3k^{2}+2}\right)}{2k^{2}+1}-k$$

Multiply k times \frac{\sqrt{2}\left(\sqrt{2}k^{2}-\sqrt{3k^{2}+2}\right)}{1+2k^{2}}.

$$y=\frac{\sqrt{2}\left(\sqrt{2}k^{2}-\sqrt{3k^{2}+2}\right)}{2k^{2}+1}k-k$$

The system is now solved.

$$y=\frac{\sqrt{2}\left(\sqrt{2}k^{2}+\sqrt{3k^{2}+2}\right)}{2k^{2}+1}k-k,x=\frac{\sqrt{2}\left(\sqrt{2}k^{2}+\sqrt{3k^{2}+2}\right)}{2k^{2}+1}\text{ or }y=\frac{\sqrt{2}\left(\sqrt{2}k^{2}-\sqrt{3k^{2}+2}\right)}{2k^{2}+1}k-k,x=\frac{\sqrt{2}\left(\sqrt{2}k^{2}-\sqrt{3k^{2}+2}\right)}{2k^{2}+1}$$

Solve for x, y (complex solution)

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