# Explain why a hyperbola with center $(m,n)$ has the parametrized curve $r(t) = (m + a cosh, n + b sinh)$ "laying" on it

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.

• The parameterization doesn’t involve $\cos$ and $\sin$. It uses the hyperbolic functions $\cosh$ and $\sinh$. The algebraic manipulations required to convert to an implicit Cartesian equation aren’t all that different from converting $(\cos t,\sin t)$ into $x^2+y^2=1$, though.
– amd
Feb 20, 2019 at 6:36

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 $$\left({x-m\over a}\right)^2-\left({y-n\over b}\right)^2=1.$$