Problem source

The equation of a hyperbola (with respect to axes which are displaced and rotated with respect to the standard axes) is 
	\[
	3y^{2}-10xy+3x^{2}+16y-16x+15=0.\tag{$\dagger$}
	\]
	By differentiating $(\dagger)$, or otherwise, show that the equation of the tangent through the point $(s,t)$ on the curve is 
	\[
	y=\left(\frac{5t-3s+8}{3t-5s+8}\right)x-\left(\frac{8t-8s+15}{3t-5s+8}\right).
	\]
	Show that the equations of the asymptote (the limiting tangents as $s\rightarrow\infty$) are 
	\[
	y=3x-4\qquad\mbox{ and }\qquad3y=x-4.
	\]
	{[}\textbf{Hint}: You will need to find a relationship between $s$ and $t$ which is valid in the limit as $s\rightarrow\infty.${]}
	Show that the angle between one asymptote and the $x$-axis is the same as the angle between the other asymptote and the $y$-axis. Deduce the slopes of the lines that bisect the angles between the asymptotes and find the equations of the axes of the hyperbola.

Solution source

\begin{align*}
&& 0 &= 3y^{2}-10xy+3x^{2}+16y-16x+15 \\
\Rightarrow && 0 &= 6y \frac{\d y}{\d x} - 10x \frac{\d y}{\d x} - 10y + 6x+ 16 \frac{\d y}{\d x } - 16 \\
&&&= \frac{\d y}{\d x} \left (6y - 10x +16 \right)  - (10y-6x+16) \\
\Rightarrow && \frac{\d y}{\d x} &= \frac{5y-3x+8}{3y-5x+8} \\
\Rightarrow && \frac{y-t}{x-s} &= \frac{5t-3s+8}{3t-5s+8} \\
&& y &= \left(\frac{5t-3s+8}{3t-5s+8}\right)x -\left(\frac{5t-3s+8}{3t-5s+8}\right)s + t \\
&&&= \left(\frac{5t-3s+8}{3t-5s+8}\right)x - \frac{(5ts-3s^2+8s)-(3t^2-5st+8t)}{3t-5s+8} \\
&&&= \left(\frac{5t-3s+8}{3t-5s+8}\right)x - \frac{(5ts-3s^2+8s)-(3t^2-5st+8t)}{3t-5s+8} \\
&&&= \left(\frac{5t-3s+8}{3t-5s+8}\right)x - \frac{8s-8t-(3s^2+3t^2-10st)}{3t-5s+8} \\ 
&&&=  \left(\frac{5t-3s+8}{3t-5s+8}\right)x - \frac{8s-8t-(16s-16t-15)}{3t-5s+8} \\ 
&&&= \left(\frac{5t-3s+8}{3t-5s+8}\right)x - \frac{8t-8s+15}{3t-5s+8} \\ 
\end{align*}

While $x \to \infty$ we still have $3 \frac{y^2}{x^2} - 10 \frac{y}{x} + 3 + 16 \frac{y}{x^2} - 16\frac{1}{x} + 15 \frac{1}{x^2} = 0$, ie if $\frac{y}{x} = k$, then $3k^2 - 10k + 3 \to 0 \Rightarrow k \to 3, \frac13$. Therefore, as $s \to \infty$ we can write

\begin{align*}
&& y &=  \left(\frac{5\frac{t}{s}-3+8\frac{1}{s}}{3\frac{t}{s}-5+8\frac1{s}}\right)x - \frac{8\frac{t}s-8+15\frac{1}{s}}{3\frac{t}{s}-5+8\frac{1}{s}} \\
k \to 3: &&& \to  \left(\frac{15-3}{9-5}\right)x - \frac{24-8}{9-5} \\
&&&= 3x - 4 \\
k \to \frac13: && &\to \left(\frac{\frac53-3}{1-5}\right)x - \frac{\frac83-8}{1-5} \\ 
&&&= \frac13 x - \frac43
\end{align*}

Therefore the equations are $y = 3x-4$ and $3y=x-4$

The lines are parallel to $y = 3x$ and $y = \frac13x$, so by considering the triangles formed with the origin and a point $1$ along the $x$ or $y$ axis we can see the angles are identical. This means the line $y = x$ is parallel to one axis and $y = -x$ is parallel to the other. They must meet where our two lines meet which is $(1,-1)$, so our lines are $y = x-2$ and $y = -x$