Solve {l}{y=k(x-a)}{5x^2+9y^2=5a}

$$\left\{ \begin{array} { l } { y = k ( x - a ) } \\ { 5 x ^ { 2 } + 9 y ^ { 2 } = 5 a } \end{array} \right.$$

Solve for x, y

$$x=\frac{9ak^{2}-\sqrt{5a\left(5+9k^{2}-9ak^{2}\right)}}{9k^{2}+5}\text{, }y=-\frac{\sqrt{5}k\left(\sqrt{a\left(5+9k^{2}-9ak^{2}\right)}+\sqrt{5}a\right)}{9k^{2}+5}$$

$$x=\frac{9ak^{2}+\sqrt{5a\left(5+9k^{2}-9ak^{2}\right)}}{9k^{2}+5}\text{, }y=\frac{k\left(\sqrt{5a\left(5+9k^{2}-9ak^{2}\right)}-5a\right)}{9k^{2}+5}\text{, }\left(a>1\text{ and }|k|\leq \frac{\sqrt{-\frac{5}{1-a}}}{3}\right)\text{ or }\left(a\geq 0\text{ and }a\leq 1\right)$$

Steps Using Substitution and Quadratic Formula

Consider the first equation. Use the distributive property to multiply k by x-a.

$$y=kx-ka$$

Subtract kx from both sides.

$$y-kx=-ka$$

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.

$$y+\left(-k\right)x=-ak,5x^{2}+9y^{2}=5a$$

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

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

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

$$y=kx-ak$$

Substitute kx-ak for y in the other equation, 5x^{2}+9y^{2}=5a.

$$5x^{2}+9\left(kx-ak\right)^{2}=5a$$

Square kx-ak.

$$5x^{2}+9\left(k^{2}x^{2}+2k\left(-ak\right)x+\left(-ak\right)^{2}\right)=5a$$

Multiply 9 times k^{2}x^{2}+2k\left(-ak\right)x+\left(-ak\right)^{2}.

$$5x^{2}+9k^{2}x^{2}+18k\left(-ak\right)x+9\left(-ak\right)^{2}=5a$$

Add 5x^{2} to 9k^{2}x^{2}.

$$\left(9k^{2}+5\right)x^{2}+18k\left(-ak\right)x+9\left(-ak\right)^{2}=5a$$

Subtract 5a from both sides of the equation.

$$\left(9k^{2}+5\right)x^{2}+18k\left(-ak\right)x+9\left(-ak\right)^{2}-5a=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5+9k^{2} for a, 9\times 2k\left(-ak\right) for b, and a\left(9ak^{2}-5\right) for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-18k\left(-ak\right)±\sqrt{\left(18k\left(-ak\right)\right)^{2}-4\left(9k^{2}+5\right)a\left(9ak^{2}-5\right)}}{2\left(9k^{2}+5\right)}$$

Square 9\times 2k\left(-ak\right).

$$x=\frac{-18k\left(-ak\right)±\sqrt{324a^{2}k^{4}-4\left(9k^{2}+5\right)a\left(9ak^{2}-5\right)}}{2\left(9k^{2}+5\right)}$$

Multiply -4 times 5+9k^{2}.

$$x=\frac{-18k\left(-ak\right)±\sqrt{324a^{2}k^{4}+\left(-36k^{2}-20\right)a\left(9ak^{2}-5\right)}}{2\left(9k^{2}+5\right)}$$

Multiply -20-36k^{2} times a\left(9ak^{2}-5\right).

$$x=\frac{-18k\left(-ak\right)±\sqrt{324a^{2}k^{4}-4a\left(9k^{2}+5\right)\left(9ak^{2}-5\right)}}{2\left(9k^{2}+5\right)}$$

Add 324a^{2}k^{4} to -4\left(5+9k^{2}\right)a\left(9ak^{2}-5\right).

$$x=\frac{-18k\left(-ak\right)±\sqrt{20a\left(5+9k^{2}-9ak^{2}\right)}}{2\left(9k^{2}+5\right)}$$

Take the square root of 20a\left(-9ak^{2}+9k^{2}+5\right).

$$x=\frac{-18k\left(-ak\right)±2\sqrt{5a\left(5+9k^{2}-9ak^{2}\right)}}{2\left(9k^{2}+5\right)}$$

Multiply 2 times 5+9k^{2}.

$$x=\frac{18ak^{2}±2\sqrt{5a\left(5+9k^{2}-9ak^{2}\right)}}{18k^{2}+10}$$

Now solve the equation x=\frac{18ak^{2}±2\sqrt{5a\left(5+9k^{2}-9ak^{2}\right)}}{18k^{2}+10} when ± is plus. Add 18ak^{2} to 2\sqrt{5a\left(-9ak^{2}+9k^{2}+5\right)}.

$$x=\frac{18ak^{2}+2\sqrt{5a\left(5+9k^{2}-9ak^{2}\right)}}{18k^{2}+10}$$

Divide 18ak^{2}+2\sqrt{5a\left(-9ak^{2}+9k^{2}+5\right)} by 10+18k^{2}.

$$x=\frac{9ak^{2}+\sqrt{5a\left(5+9k^{2}-9ak^{2}\right)}}{9k^{2}+5}$$

Now solve the equation x=\frac{18ak^{2}±2\sqrt{5a\left(5+9k^{2}-9ak^{2}\right)}}{18k^{2}+10} when ± is minus. Subtract 2\sqrt{5a\left(-9ak^{2}+9k^{2}+5\right)} from 18ak^{2}.

$$x=\frac{18ak^{2}-2\sqrt{5a\left(5+9k^{2}-9ak^{2}\right)}}{18k^{2}+10}$$

Divide 18ak^{2}-2\sqrt{5a\left(-9ak^{2}+9k^{2}+5\right)} by 10+18k^{2}.

$$x=\frac{9ak^{2}-\sqrt{5a\left(5+9k^{2}-9ak^{2}\right)}}{9k^{2}+5}$$

There are two solutions for x: \frac{9ak^{2}+\sqrt{5a\left(-9ak^{2}+9k^{2}+5\right)}}{5+9k^{2}} and \frac{9ak^{2}-\sqrt{5a\left(-9ak^{2}+9k^{2}+5\right)}}{5+9k^{2}}. Substitute \frac{9ak^{2}+\sqrt{5a\left(-9ak^{2}+9k^{2}+5\right)}}{5+9k^{2}} for x in the equation y=kx-ak to find the corresponding solution for y that satisfies both equations.

$$y=k\times \frac{9ak^{2}+\sqrt{5a\left(5+9k^{2}-9ak^{2}\right)}}{9k^{2}+5}-ak$$

Now substitute \frac{9ak^{2}-\sqrt{5a\left(-9ak^{2}+9k^{2}+5\right)}}{5+9k^{2}} for x in the equation y=kx-ak and solve to find the corresponding solution for y that satisfies both equations.

$$y=k\times \frac{9ak^{2}-\sqrt{5a\left(5+9k^{2}-9ak^{2}\right)}}{9k^{2}+5}-ak$$

The system is now solved.

$$y=k\times \frac{9ak^{2}+\sqrt{5a\left(5+9k^{2}-9ak^{2}\right)}}{9k^{2}+5}-ak,x=\frac{9ak^{2}+\sqrt{5a\left(5+9k^{2}-9ak^{2}\right)}}{9k^{2}+5}\text{ or }y=k\times \frac{9ak^{2}-\sqrt{5a\left(5+9k^{2}-9ak^{2}\right)}}{9k^{2}+5}-ak,x=\frac{9ak^{2}-\sqrt{5a\left(5+9k^{2}-9ak^{2}\right)}}{9k^{2}+5}$$

Solve for x, y (complex solution)

$$\left\{\begin{matrix}x=\frac{9ak^{2}-\sqrt{5a\left(5+9k^{2}-9ak^{2}\right)}}{9k^{2}+5}\text{, }y=-\frac{\sqrt{5}k\left(\sqrt{a\left(5+9k^{2}-9ak^{2}\right)}+\sqrt{5}a\right)}{9k^{2}+5}\text{; }x=\frac{9ak^{2}+\sqrt{5a\left(5+9k^{2}-9ak^{2}\right)}}{9k^{2}+5}\text{, }y=\frac{k\left(\sqrt{5a\left(5+9k^{2}-9ak^{2}\right)}-5a\right)}{9k^{2}+5}\text{, }&k\neq -\frac{\sqrt{5}i}{3}\text{ and }k\neq \frac{\sqrt{5}i}{3}\\x=\frac{9ak^{2}-5}{18k^{2}}\text{, }y=\frac{-9ak^{2}-5}{18k}\text{, }&a\neq 0\text{ and }\left(k=-\frac{\sqrt{5}i}{3}\text{ or }k=\frac{\sqrt{5}i}{3}\right)\\x\in \mathrm{C}\text{, }y=k\left(x-a\right)\text{, }&\left(k=\frac{\sqrt{5}i}{3}\text{ or }k=-\frac{\sqrt{5}i}{3}\right)\text{ and }a=0\end{matrix}\right.$$