Proof that the hyperbolae $x^2-y^2 =a$ and $xy=b$ are orthogonal via parametric equations 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.
 A: 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.
A: There is no need for parametric equations. Take an arbitrary point $(x,y)$. By differentiating the implicit equations,
$$\begin{cases}x-yy_a'=0,\\ y+xy_b'=0\end{cases}$$
implies $$y_a'y_b'=-1.$$

If you really want parametric equations,
$$(x+y)(x-y)=a\iff x+y=t,x-y=\frac at\iff x=\frac12\left(t+\frac at\right),y=\frac12\left(t-\frac at\right).$$
$$xy=b\iff x=u,y=\frac bu.$$
The tangent vectors are proportional to
$$\frac12\left(1-\frac a{t^2},1+\frac a{t^2}\right)=\frac1t(y,x)$$
and
$$\left(1,-\frac b{u^2}\right)=\frac1u(x,-y).$$
Obviously, the dot product is null. 
