Struggling with how to approach a task given at university.
The first part of the task asked to show that the equation $$4x^2-32x-9y^2+36y=8$$ resulted in a hyperbola, and to find the center, asymptotes, etc.
From my (hopefully correct) calculations the hyperbola can be expressed on the form $${(x-m)^2\over a^2} - {(y-n)^2\over b^2} = 1$$ where the center is expressed by $(m,n)$, where $m=4$ and $n=2$. The semi major axis i found to be $a=3$, and $b=2$
Assuming I haven't made any glaring mistakes, this is where I get stuck. The second part of the task asks that given the same center $(m,n)$ and values for $a$ and $b$, explain how the parametrized curve $$r(t)=(m+a cosh(t), n+ bsinh(t))$$ "lays on the hyperbola". The hints given are that $$cosh(x)={1\over2}(e^x-e^{-x})$$ $$sinh(x)={1\over2}(e^x-e^{-x})$$ and and that $$cosh^2(x) - sinh^2 (x)=1$$.
Just looking at it I can sort of make out that the parametrized curve has coordinates based on the center $(m,n)$ and that you essentially just plot the coordinates based on a transformation, but I can't work out how to arrive at this conclusion, or how to explain it.
Might have something to do with the fact that I am generally pretty bad at grasping functions of $cos$ and $sin$. It's a work in progress, so please go easy on me. Basically just gotten started with multivariable calculus and linear algebra if that helps with formulating answers.
All you really need is the last identity. From $$(x,y) = (m+a\cosh t,n+b\sinh t)$$ you have $$\cosh t = {x-m\over a} \\ \sinh t = {y-n\over b}$$ and substituting into that equality results in $$\left({x-m\over a}\right)^2-\left({y-n\over b}\right)^2=1.$$