Question 10.4.5: Rotation of axes for a hyperbola Use rotation of axes to wri...
Rotation of axes for a hyperbola
Use rotation of axes to write y = 1/x in the standard form for a hyperbola.
Since y = 1/x is equivalent to xy – 1 = 0, we have A = 0, B = 1, and C = 0. Use \cot 2 \theta=\frac{A-C}{B} to get \cot 2 \theta=0 or \theta=\pi / 4. Now find x and y in terms of x’ and y’:
Substitute these expressions for x and y into xy – 1 = 0:
\begin{aligned} \left(\frac{x^{\prime}-y^{\prime}}{\sqrt{2}}\right)\left(\frac{x^{\prime}+y^{\prime}}{\sqrt{2}}\right)-1 &=0 \\ \frac{\left(x^{\prime}\right)^{2}-\left(y^{\prime}\right)^{2}}{2}-1 &=0 \\ \frac{\left(x^{\prime}\right)^{2}}{2}-\frac{\left(y^{\prime}\right)^{2}}{2} &=1 \end{aligned}In the x’y’-coordinate system, \frac{\left(x^{\prime} \right)^{2}}{2}-\frac{\left(y^{\prime}\right)^{2}}{2}=1 is the equation of a hyperbola centered at the origin with vertices at (\pm \sqrt{2}, 0), as shown in Fig. 10.45. The asymptotes are y^{\prime}=\pm x^{\prime}, which are the original x– and y-axes. To find the xy-coordinates of the vertices, let x^{\prime}=\pm \sqrt{2} and y’ = 0 in the equations
x=\frac{x^{\prime}-y^{\prime}}{\sqrt{2}} and y=\frac{x^{\prime}+y^{\prime}}{\sqrt{2}}
to get the vertices (1, 1) and (-1, -1) in the xy-coordinates.
