Hallo Bertie,
i have finished your wonderful solution.
\(\begin{array}{rcll} \text{I set slope line 1: } ~ m_1 = \frac{13}{10} \\ \text{and slope line 2: } ~ m_2 = \frac{5}{6} \\ \text{and your equation: }~Y^2 + aX^2 + bXY + c &=& 0. \qquad | \quad X = x - \frac{32}{7},~ Y = y - \frac87\\ \text{and your equations of the two asymptotes: }~(\frac{Y}{X})^2 + b\cdot\frac{Y}{X} + a &=& 0\\ \text{and with solution of a quadratic: }~\frac{Y}{X} &=& \frac12 \cdot (-b \pm \sqrt{b^2 - 4a})\\ \text{we can find a and b: } \end{array}\)
\(\begin{array}{rcll} \frac{Y}{X} &=& \frac12 \cdot (-b \pm \sqrt{b^2 - 4a}) \\ 2\cdot \frac{Y}{X}+b &=& \pm \sqrt{b^2 - 4a} \qquad | \qquad \text{(square both sides)} \\ (2\cdot \frac{Y}{X}+b)^2 &=& b^2 - 4a \\ 4\cdot (\frac{Y}{X})^2 +4\cdot\frac{Y}{X}\cdot b +b^2 &=& b^2 - 4a \\ 4\cdot (\frac{Y}{X})^2 +4\cdot\frac{Y}{X}\cdot b &=& - 4a \\ (\frac{Y}{X})^2 + \frac{Y}{X}b &=& -a \qquad | \qquad \text{we set }~ \frac{Y}{X}=m_1 \text{ and }\frac{Y}{X}=m_2\\ m_1^2+m_1b =&-a& =m_2^2+m_2b \\ b\cdot (m_1-m_2) &=& m_2^2-m_1^2 \\ b\cdot (m_1-m_2) &=&(m_2-m_1)(m_1+m_2) \\ b\cdot (m_1-m_2) &=&-(m_1-m_2)(m_1+m_2) \\ \mathbf{b} & \mathbf{=} & \mathbf{-(m_1+m_2)} = - (\frac{13}{10}+\frac{5}{6})=-\frac{32}{15} \\\\ m_1^2+m_1b &=& -a \\ m_2^2+m_2b &=& -a \\ b&=& \frac{-a-m_1^2}{m_1} = \frac{-a-m_2^2}{m_2} \\ m_2( -a-m_1^2 ) &=& m_1(-a-m_2^2 )\\ \cdots \\ a\cdot (m_1-m_2) &=& m_1\cdot m_2\cdot (m_1-m_2) \\ \mathbf{a} & \mathbf{=} & \mathbf{m_1\cdot m_2} = \frac{13}{10}\cdot \frac{5}{6} = \frac{13}{12}\\ \end{array}\)
\(\begin{array}{lcll} \text{your equation is now: } \end{array}\\ \begin{array}{rcll} Y^2 + m_1\cdot m_2\cdot X^2 - (m_1+m_2)\cdot XY + c &=& 0. \qquad | \quad X = x - \frac{32}{7},~ Y = y - \frac87\\ \end{array}\\ \begin{array}{lcll} \text{or: } \end{array}\\ \begin{array}{rcll} \mathbf{(y-\frac87)^2 + \frac{13}{10}\cdot \frac{5}{6} \cdot (x-\frac{32}{7})^2 - (\frac{13}{10}+\frac{5}{6})\cdot (x-\frac{32}{7}) \cdot (y-\frac87)+c }&\mathbf{=}& \mathbf{0} \end{array}\)
You just asked a question bro.
