Solve {l}{x^2/4+y^2/3=1}{y=k(x-sqrt{2})}

$$\left\{ \begin{array} { l } { \frac { x ^ { 2 } } { 4 } + \frac { y ^ { 2 } } { 3 } = 1 } \\ { y = k ( x - \sqrt { 2 } ) } \end{array} \right.$$

Solve for x, y

$$x=\frac{2\left(2\sqrt{2}k^{2}-\sqrt{3\left(2k^{2}+3\right)}\right)}{4k^{2}+3}\text{, }y=-\frac{\sqrt{2}k\left(\sqrt{6\left(2k^{2}+3\right)}+3\right)}{4k^{2}+3}$$

$$x=\frac{2\left(2\sqrt{2}k^{2}+\sqrt{3\left(2k^{2}+3\right)}\right)}{4k^{2}+3}\text{, }y=\frac{\sqrt{2}k\left(\sqrt{6\left(2k^{2}+3\right)}-3\right)}{4k^{2}+3}$$

Steps Using Substitution and Quadratic Formula

Consider the first equation. Multiply both sides of the equation by 12, the least common multiple of 4,3.

$$3x^{2}+4y^{2}=12$$

Consider the second equation. Use the distributive property to multiply k by x-\sqrt{2}.

$$y=kx-k\sqrt{2}$$

Subtract kx from both sides.

$$y-kx=-k\sqrt{2}$$

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.

$$y+\left(-k\right)x=-\sqrt{2}k,3x^{2}+4y^{2}=12$$

Solve y+\left(-k\right)x=-\sqrt{2}k for y by isolating y on the left hand side of the equal sign.

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

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

$$y=kx-\sqrt{2}k$$

Substitute kx-\sqrt{2}k for y in the other equation, 3x^{2}+4y^{2}=12.

$$3x^{2}+4\left(kx-\sqrt{2}k\right)^{2}=12$$

Square kx-\sqrt{2}k.

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

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

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

Add 3x^{2} to 4k^{2}x^{2}.

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

Subtract 12 from both sides of the equation.

$$\left(4k^{2}+3\right)x^{2}+8k\left(-\sqrt{2}k\right)x+4\left(-\sqrt{2}k\right)^{2}-12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3+4k^{2} for a, 4\times 2k\left(-k\sqrt{2}\right) for b, and 8k^{2}-12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

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

$$x=\frac{-8k\left(-\sqrt{2}k\right)±\sqrt{128k^{4}-4\left(4k^{2}+3\right)\left(8k^{2}-12\right)}}{2\left(4k^{2}+3\right)}$$

Multiply -4 times 3+4k^{2}.

$$x=\frac{-8k\left(-\sqrt{2}k\right)±\sqrt{128k^{4}+\left(-16k^{2}-12\right)\left(8k^{2}-12\right)}}{2\left(4k^{2}+3\right)}$$

Multiply -12-16k^{2} times 8k^{2}-12.

$$x=\frac{-8k\left(-\sqrt{2}k\right)±\sqrt{128k^{4}+144+96k^{2}-128k^{4}}}{2\left(4k^{2}+3\right)}$$

Add 128k^{4} to 144+96k^{2}-128k^{4}.

$$x=\frac{-8k\left(-\sqrt{2}k\right)±\sqrt{96k^{2}+144}}{2\left(4k^{2}+3\right)}$$

Take the square root of 96k^{2}+144.

$$x=\frac{-8k\left(-\sqrt{2}k\right)±4\sqrt{6k^{2}+9}}{2\left(4k^{2}+3\right)}$$

Multiply 2 times 3+4k^{2}.

$$x=\frac{8\sqrt{2}k^{2}±4\sqrt{6k^{2}+9}}{8k^{2}+6}$$

Now solve the equation x=\frac{8\sqrt{2}k^{2}±4\sqrt{6k^{2}+9}}{8k^{2}+6} when ± is plus. Add 8k^{2}\sqrt{2} to 4\sqrt{6k^{2}+9}.

$$x=\frac{8\sqrt{2}k^{2}+4\sqrt{6k^{2}+9}}{8k^{2}+6}$$

Divide 8\sqrt{2}k^{2}+4\sqrt{6k^{2}+9} by 6+8k^{2}.

$$x=\frac{2\left(2\sqrt{2}k^{2}+\sqrt{6k^{2}+9}\right)}{4k^{2}+3}$$

Now solve the equation x=\frac{8\sqrt{2}k^{2}±4\sqrt{6k^{2}+9}}{8k^{2}+6} when ± is minus. Subtract 4\sqrt{6k^{2}+9} from 8k^{2}\sqrt{2}.

$$x=\frac{8\sqrt{2}k^{2}-4\sqrt{6k^{2}+9}}{8k^{2}+6}$$

Divide 8\sqrt{2}k^{2}-4\sqrt{6k^{2}+9} by 6+8k^{2}.

$$x=\frac{2\left(2\sqrt{2}k^{2}-\sqrt{6k^{2}+9}\right)}{4k^{2}+3}$$

There are two solutions for x: \frac{2\left(2k^{2}\sqrt{2}+\sqrt{9+6k^{2}}\right)}{3+4k^{2}} and \frac{2\left(2k^{2}\sqrt{2}-\sqrt{9+6k^{2}}\right)}{3+4k^{2}}. Substitute \frac{2\left(2k^{2}\sqrt{2}+\sqrt{9+6k^{2}}\right)}{3+4k^{2}} for x in the equation y=kx-\sqrt{2}k to find the corresponding solution for y that satisfies both equations.

$$y=k\times \frac{2\left(2\sqrt{2}k^{2}+\sqrt{6k^{2}+9}\right)}{4k^{2}+3}-\sqrt{2}k$$

Multiply k times \frac{2\left(2k^{2}\sqrt{2}+\sqrt{9+6k^{2}}\right)}{3+4k^{2}}.

$$y=\frac{2\left(2\sqrt{2}k^{2}+\sqrt{6k^{2}+9}\right)}{4k^{2}+3}k-\sqrt{2}k$$

Now substitute \frac{2\left(2k^{2}\sqrt{2}-\sqrt{9+6k^{2}}\right)}{3+4k^{2}} for x in the equation y=kx-\sqrt{2}k and solve to find the corresponding solution for y that satisfies both equations.

$$y=k\times \frac{2\left(2\sqrt{2}k^{2}-\sqrt{6k^{2}+9}\right)}{4k^{2}+3}-\sqrt{2}k$$

Multiply k times \frac{2\left(2k^{2}\sqrt{2}-\sqrt{9+6k^{2}}\right)}{3+4k^{2}}.

$$y=\frac{2\left(2\sqrt{2}k^{2}-\sqrt{6k^{2}+9}\right)}{4k^{2}+3}k-\sqrt{2}k$$

The system is now solved.

$$y=\frac{2\left(2\sqrt{2}k^{2}+\sqrt{6k^{2}+9}\right)}{4k^{2}+3}k-\sqrt{2}k,x=\frac{2\left(2\sqrt{2}k^{2}+\sqrt{6k^{2}+9}\right)}{4k^{2}+3}\text{ or }y=\frac{2\left(2\sqrt{2}k^{2}-\sqrt{6k^{2}+9}\right)}{4k^{2}+3}k-\sqrt{2}k,x=\frac{2\left(2\sqrt{2}k^{2}-\sqrt{6k^{2}+9}\right)}{4k^{2}+3}$$

Solve for x, y (complex solution)

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