hyperbola equation to standard/canonical equation

Question

I have this hyperbola in the Cartesian plane.

$$f(x) = y = \frac{x+1}{2x-3}$$

I want to get it to its standard/canonical form:

https://en.wikipedia.org/wiki/Hyperbola#Hyperbola_in_Cartesian_coordinates

$$\frac{x_1^2}{a^2} - \frac{y_1^2}{b^2} = 1$$

For this, I guess I need to change the coordinate system $Oxy$ to another Cartesian one $Ox_1y_1$

How do I find the change/transformation equations from $(x,y)$ to $(x_1, y_1)$?

I used to know this from analytic geometry but I have forgotten it, it seems.

Answer 1

$y=\frac{x+1}{2x-3}$ or $2xy-3y=x+1$ or $2(x-\frac32)(y-\frac12)-\frac52=0.$

You can find the center by solving $\frac{\partial}{\partial x} (2xy-3y-x-1)=0,\frac{\partial}{\partial y} (2xy-3y-x-1)=0.$

Expanding $2(X+\frac32)(Y+\frac12)-3(Y+\frac12)-(X+\frac32)-1=0$ gives $2XY-\frac52=0,$ now $x=X+\frac32$ or $X=x-\frac32$ and similarly for $Y.$

Now rotate by $\frac{\pi}{4}$:

See this answer to see that angles to rotate by are $\theta$ such that $\cos{2\theta}=0$

$$2\frac{x'+y'}{\sqrt{2}}\frac{x'-y'}{\sqrt{2}}=\frac52$$ or $$x'^2-y'^2=\frac52$$

Remember $X=\frac{x'+y'}{\sqrt{2}},Y=\frac{x'-y'}{\sqrt{2}},$ and this transformation is its own inverse so $x'=\frac{X+Y}{\sqrt{2}},y'=\frac{X-Y}{\sqrt{2}},$ making $x'=\frac{(x-\frac32)+(y-\frac12)}{\sqrt{2}},y'=\frac{(x-\frac32)-(y-\frac12)}{\sqrt{2}}.$

or $$\frac52(\frac{(\frac{(x-\frac32)+(y-\frac12)}{\sqrt{2}})^2}{\frac52}-\frac{(\frac{(x-\frac32)-(y-\frac12)}{\sqrt{2}})^2}{\frac52}-1)=0.$$