# Find the equations of the hyperbolas that intersect 3x^2-4y^2=5xy and 3y^2-4x^2=2x+5?

## Find the equations of the hyperbolas that intersect $3 {x}^{2} - 4 {y}^{2} = 5 x y$ and $3 {y}^{2} - 4 {x}^{2} = 2 x + 5$

Mar 22, 2017

Given:

$\left\{\begin{matrix}3 {x}^{2} - 4 {y}^{2} = 5 x y \\ 3 {y}^{2} - 4 {x}^{2} = 2 x + 5\end{matrix}\right.$

Note that the first of these "hyperbolas" is degenerate, being the union of two straight lines.

graph{(3y^2-4x^2-2x-5)(3x^2-4y^2-5xy) = 0 [-6, 6, -3, 3]}

Subtracting $5 x y$ from both sides of the first equation, we get:

$3 {x}^{2} - 5 x y - 4 {y}^{2} = 0$

Multiply through by $12$, then complete the square as follows:

$0 = 12 \left(3 {x}^{2} - 5 x y - 4 {y}^{2}\right)$

$\textcolor{w h i t e}{0} = 36 {x}^{2} - 60 x y - 48 {y}^{2}$

$\textcolor{w h i t e}{0} = {\left(6 x\right)}^{2} - 2 \left(6 x\right) \left(5 y\right) + {\left(5 y\right)}^{2} - 73 {y}^{2}$

$\textcolor{w h i t e}{0} = {\left(6 x - 5 y\right)}^{2} - {\left(\sqrt{73} y\right)}^{2}$

$\textcolor{w h i t e}{0} = \left(\left(6 x - 5 y\right) - \sqrt{73} y\right) \left(\left(6 x - 5 y\right) + \sqrt{73} y\right)$

$\textcolor{w h i t e}{0} = \left(6 x - \left(5 + \sqrt{73}\right) y\right) \left(6 x - \left(5 - \sqrt{73}\right) y\right)$

The other (proper) hyperbola intersects the line:

$y = \frac{6}{5 - \sqrt{73}} x$

Substitute this expression for $y$ into the second equation to get:

$3 {\left(\frac{6}{5 - \sqrt{73}} x\right)}^{2} - 4 {x}^{2} = 2 x + 5$

That is:

$\left(4 - 3 {\left(\frac{6}{5 - \sqrt{73}}\right)}^{2}\right) {x}^{2} + 2 x + 5 = 0$

Hence:

$x = - \frac{4 \left(\sqrt{254 + 150 \sqrt{73}} \pm 8\right)}{19 + 15 \sqrt{73}}$

Then:

$y = \frac{6}{5 - \sqrt{73}} x$