could anyone help me with this problem?
Prove that the hyperbolae $x^2-y^2 =a$ and $xy=b$ are orthogonal to each other at each point they intersect. Here $a$ and $b$ are non zero parameters.
I first did a parameterization
$$y=t$$ $$x=\frac{b}{t}$$
Then I differentiated both equations to get the unit tangent vector. $$y'=1$$ $$x'=\frac{-b}{t^2}$$
From here onward I'm unsure of how to proceed.
Implicit derivatives will work nicely here. First we assume $b\neq 0$ or else the second hyperbola really is just the two axes (not a hyperbola). Then any point $(x,y)$ on both curves at once must have $x,y \neq 0.$ Implicit differentiation of $x^2-y^2=a$ gives $2x-2yy'=0,\ y'=x/y.$ on the other hand, implicit differentiation of $xy=b$ gives $xy'+y=0,\ y'=-y/x.$ Thus at any point on both hyperbolas, one slope is $x/y$ while the other is its negative reciprocal $-y/x$, showing the curves are orthogonal where they meet.