Find the values of a and b that make the hyperbola congruent to $xy=1$ I need to find a and b that would make the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ congruent to the rectangular hyperbola $xy=1$.
I know that the answer is $a=b=\sqrt{2}$, and I've found some answers that prove it (using polar coordinates), however, I haven't found anything on the process of how to actually find a and b. Furthermore, we haven't gone over polar coordinates yet so I need an answer that doesn't involve that.
I've been searching for over an hour both on here and other websites and I can't seem to find what I'm looking for.
 A: You should know that a hyperbola $H_1$ with equation $xy=1$ has the coordinate axes as asymptotes. Hence its asymptotes are perpendicular.
A hyperbola $H_2$ with equation ${x^2\over a^2}-{y^2\over b^2}=1$ has two asymptotes with slopes $\pm{b\over a}$. They are perpendicular only if $a=b$.
To find the value of $a$, take for instance a vertex $V_1=(1,1)$ in $H_1$: its coordinates are the distances of $V$ from the asymptotes. The corresponding point in $H_2$ is $V_2=(a,0)$ and its distances from the asymptotes (which are lines $y=\pm x$) are both $a/\sqrt2$. A comparison with $V_1$ gives $a/\sqrt2=1$, i.e. $a=\sqrt2$.
A: Rotate the x-y coordinate system  CW by $45^\circ$ to x'-y' system, we have the following transformation   equations:
\begin{align} x=\frac{x'-y'}{\sqrt 2}, y=\frac{x'+y'}{\sqrt 2}  \end{align}
Thus,
\begin{align} x y=(\frac{x'+y'}{\sqrt 2})(\frac{x'-y'}{\sqrt 2})=  (\frac{x'^2}{\sqrt 2^2})-(\frac{y'^2}{\sqrt 2^2})=1, a=b=\sqrt 2 \end{align}

A: Put $y=0$. Its  x- axis tendency/ behavior is noted to be different. So without some transformation there may be no direct congruence.
By rotation of $x-,y-$ axes it is noted that quadratic term coefficients and ( same exponents 2) can be compared. Axes rotation is by $45^{\circ}$
$$ x\to (x_1-y_1)/ \sqrt 2  \; ; y \to (x_1+y_1)/ \sqrt 2 ;$$
Plug into the second relation
$$\dfrac{x_1^2}{2}-\dfrac{y_1^2}{2}=1 $$
Compare coefficients of given ellipse equation  ( drop suffixes for rotated axes)
$$\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1 $$
to get the answer.
A: Note that the hyperbola $xy=1$ is symmetric with respect to the line $y=x$, which indicates that its vertexes are $(\pm 1,\pm 1)$ and its major semi-axis is $\sqrt{1^2 + 1^2}=\sqrt2$. Moreover, the $x$- and $y$-axes are its asymptotes, implying a square hyperbola. Thus, $a=b =\sqrt2$.
