Prove to be hyperbolae $x^2 - y^2 = a$ and $xy = b$ intersect at right angle.

Question

Prove to be hyperbolae $x^2 - y^2 = a$ and $xy = b$ intersect at right angle.
My idea: $$h_1:=x^2 - y^2 = a$$ $$h_2:=xy = b$$ By using implicit differentiation we can find $h_1'$ and $h_2'$. $$x^2 - y^2 = a$$ $$2x-2yy'=0$$ $$y'=\frac{x}{y} \tag{1}$$ $$xy = b$$ $$y+xy'=0$$ $$y'=-\frac{y}{x} \tag{2} $$

This prove that $h_1$ and $h_2$ intersect at right angle for all $x$ and $y$, but $x$ any $y$ can't be equal to zero.
Is my work ok?

Answer 1

Let $(X,Y)$ be the intersection, then the equations of the tangents of the conics are:

$$ \left \{ \begin{eqnarray*} Xx-Yy &=& a \\ \frac{Yx+Xy}{2} &=& b \end{eqnarray*} \right. $$

Their slopes are $\displaystyle \frac{X}{Y}$ and $\displaystyle -\frac{Y}{X}$ respectively.

Hence they are perpendicular.

Answer 2

Note that slopes of (1), (2) (m1,m2) have a product -1, a condition of orthogonality..it is done. Because in

$$ \tan \pi/2 = \infty = \frac{m1-m2}{1+ m1 \,m2} $$

denominator is zero

Answer 3

Your solution looks good to me.

You can also do this using gradients instead of implicit differentiation: $$ \nabla h_1 \cdot \nabla h_2 = (2x,-2y)\cdot(y,x) = 0. $$ The gradients are normal to the level curves, so since they are orthogonal, then so are the curves.