Solve {l}{y=k(x-sqrt{3})}{x^2+4y^2-4=0}

$$\left. \begin{array} { l } { y = k ( x - \sqrt { 3 } ) } \\ { x ^ { 2 } + 4 y ^ { 2 } - 4 = 0 } \end{array} \right.$$

Solve for x, y

$$x=\frac{2\left(2\sqrt{3}k^{2}-\sqrt{k^{2}+1}\right)}{4k^{2}+1}\text{, }y=-\frac{k\left(2\sqrt{k^{2}+1}+\sqrt{3}\right)}{4k^{2}+1}$$

$$x=\frac{2\left(2\sqrt{3}k^{2}+\sqrt{k^{2}+1}\right)}{4k^{2}+1}\text{, }y=\frac{k\left(2\sqrt{k^{2}+1}-\sqrt{3}\right)}{4k^{2}+1}$$

Steps Using Substitution and Quadratic Formula

Consider the first equation. Use the distributive property to multiply k by x-\sqrt{3}.

$$y=kx-k\sqrt{3}$$

Subtract kx from both sides.

$$y-kx=-k\sqrt{3}$$

Consider the second equation. Add 4 to both sides. Anything plus zero gives itself.

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

To solve a pair of equations using substitution, first solve one of the equations for one of the variables. Then substitute the result for that variable in the other equation.

$$y+\left(-k\right)x=-\sqrt{3}k,x^{2}+4y^{2}=4$$

Solve y+\left(-k\right)x=-\sqrt{3}k for y by isolating y on the left hand side of the equal sign.

$$y+\left(-k\right)x=-\sqrt{3}k$$

Subtract \left(-k\right)x from both sides of the equation.

$$y=kx-\sqrt{3}k$$

Substitute kx-\sqrt{3}k for y in the other equation, x^{2}+4y^{2}=4.

$$x^{2}+4\left(kx-\sqrt{3}k\right)^{2}=4$$

Square kx-\sqrt{3}k.

$$x^{2}+4\left(k^{2}x^{2}+2k\left(-\sqrt{3}k\right)x+\left(-\sqrt{3}k\right)^{2}\right)=4$$

Multiply 4 times k^{2}x^{2}+2k\left(-\sqrt{3}k\right)x+\left(-\sqrt{3}k\right)^{2}.

$$x^{2}+4k^{2}x^{2}+8k\left(-\sqrt{3}k\right)x+4\left(-\sqrt{3}k\right)^{2}=4$$

Add x^{2} to 4k^{2}x^{2}.

$$\left(4k^{2}+1\right)x^{2}+8k\left(-\sqrt{3}k\right)x+4\left(-\sqrt{3}k\right)^{2}=4$$

Subtract 4 from both sides of the equation.

$$\left(4k^{2}+1\right)x^{2}+8k\left(-\sqrt{3}k\right)x+4\left(-\sqrt{3}k\right)^{2}-4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+4k^{2} for a, 4\times 2k\left(-k\sqrt{3}\right) for b, and 12k^{2}-4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-8k\left(-\sqrt{3}k\right)±\sqrt{\left(8k\left(-\sqrt{3}k\right)\right)^{2}-4\left(4k^{2}+1\right)\left(12k^{2}-4\right)}}{2\left(4k^{2}+1\right)}$$

Square 4\times 2k\left(-k\sqrt{3}\right).

$$x=\frac{-8k\left(-\sqrt{3}k\right)±\sqrt{192k^{4}-4\left(4k^{2}+1\right)\left(12k^{2}-4\right)}}{2\left(4k^{2}+1\right)}$$

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

$$x=\frac{-8k\left(-\sqrt{3}k\right)±\sqrt{192k^{4}+\left(-16k^{2}-4\right)\left(12k^{2}-4\right)}}{2\left(4k^{2}+1\right)}$$

Multiply -4-16k^{2} times 12k^{2}-4.

$$x=\frac{-8k\left(-\sqrt{3}k\right)±\sqrt{192k^{4}+16+16k^{2}-192k^{4}}}{2\left(4k^{2}+1\right)}$$

Add 192k^{4} to 16k^{2}-192k^{4}+16.

$$x=\frac{-8k\left(-\sqrt{3}k\right)±\sqrt{16k^{2}+16}}{2\left(4k^{2}+1\right)}$$

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

$$x=\frac{-8k\left(-\sqrt{3}k\right)±4\sqrt{k^{2}+1}}{2\left(4k^{2}+1\right)}$$

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

$$x=\frac{8\sqrt{3}k^{2}±4\sqrt{k^{2}+1}}{8k^{2}+2}$$

Now solve the equation x=\frac{8\sqrt{3}k^{2}±4\sqrt{k^{2}+1}}{8k^{2}+2} when ± is plus. Add 8\sqrt{3}k^{2} to 4\sqrt{k^{2}+1}.

$$x=\frac{8\sqrt{3}k^{2}+4\sqrt{k^{2}+1}}{8k^{2}+2}$$

Divide 8\sqrt{3}k^{2}+4\sqrt{k^{2}+1} by 2+8k^{2}.

$$x=\frac{2\left(2\sqrt{3}k^{2}+\sqrt{k^{2}+1}\right)}{4k^{2}+1}$$

Now solve the equation x=\frac{8\sqrt{3}k^{2}±4\sqrt{k^{2}+1}}{8k^{2}+2} when ± is minus. Subtract 4\sqrt{k^{2}+1} from 8\sqrt{3}k^{2}.

$$x=\frac{8\sqrt{3}k^{2}-4\sqrt{k^{2}+1}}{8k^{2}+2}$$

Divide 8\sqrt{3}k^{2}-4\sqrt{k^{2}+1} by 2+8k^{2}.

$$x=\frac{2\left(2\sqrt{3}k^{2}-\sqrt{k^{2}+1}\right)}{4k^{2}+1}$$

There are two solutions for x: \frac{2\left(2k^{2}\sqrt{3}+\sqrt{1+k^{2}}\right)}{1+4k^{2}} and \frac{2\left(2k^{2}\sqrt{3}-\sqrt{1+k^{2}}\right)}{1+4k^{2}}. Substitute \frac{2\left(2k^{2}\sqrt{3}+\sqrt{1+k^{2}}\right)}{1+4k^{2}} for x in the equation y=kx-\sqrt{3}k to find the corresponding solution for y that satisfies both equations.

$$y=k\times \frac{2\left(2\sqrt{3}k^{2}+\sqrt{k^{2}+1}\right)}{4k^{2}+1}-\sqrt{3}k$$

Multiply k times \frac{2\left(2k^{2}\sqrt{3}+\sqrt{1+k^{2}}\right)}{1+4k^{2}}.

$$y=\frac{2\left(2\sqrt{3}k^{2}+\sqrt{k^{2}+1}\right)}{4k^{2}+1}k-\sqrt{3}k$$

Now substitute \frac{2\left(2k^{2}\sqrt{3}-\sqrt{1+k^{2}}\right)}{1+4k^{2}} for x in the equation y=kx-\sqrt{3}k and solve to find the corresponding solution for y that satisfies both equations.

$$y=k\times \frac{2\left(2\sqrt{3}k^{2}-\sqrt{k^{2}+1}\right)}{4k^{2}+1}-\sqrt{3}k$$

Multiply k times \frac{2\left(2k^{2}\sqrt{3}-\sqrt{1+k^{2}}\right)}{1+4k^{2}}.

$$y=\frac{2\left(2\sqrt{3}k^{2}-\sqrt{k^{2}+1}\right)}{4k^{2}+1}k-\sqrt{3}k$$

The system is now solved.

$$y=\frac{2\left(2\sqrt{3}k^{2}+\sqrt{k^{2}+1}\right)}{4k^{2}+1}k-\sqrt{3}k,x=\frac{2\left(2\sqrt{3}k^{2}+\sqrt{k^{2}+1}\right)}{4k^{2}+1}\text{ or }y=\frac{2\left(2\sqrt{3}k^{2}-\sqrt{k^{2}+1}\right)}{4k^{2}+1}k-\sqrt{3}k,x=\frac{2\left(2\sqrt{3}k^{2}-\sqrt{k^{2}+1}\right)}{4k^{2}+1}$$

Solve for x, y (complex solution)

$$\left\{\begin{matrix}x=\frac{2\left(2\sqrt{3}k^{2}-\sqrt{k^{2}+1}\right)}{4k^{2}+1}\text{, }y=-\frac{k\left(2\sqrt{k^{2}+1}+\sqrt{3}\right)}{4k^{2}+1}\text{; }x=\frac{2\left(2\sqrt{3}k^{2}+\sqrt{k^{2}+1}\right)}{4k^{2}+1}\text{, }y=\frac{k\left(2\sqrt{k^{2}+1}-\sqrt{3}\right)}{4k^{2}+1}\text{, }&k\neq -\frac{1}{2}i\text{ and }k\neq \frac{1}{2}i\\x=\frac{\sqrt{3}\left(3k^{2}-1\right)}{6k^{2}}\text{, }y=-\frac{\sqrt{3}\left(3k^{2}+1\right)}{6k}\text{, }&k=-\frac{1}{2}i\text{ or }k=\frac{1}{2}i\end{matrix}\right.$$