Chapter A5.3, Problem 35E

(a)

To determine

To find: The new equation .

Answer to Problem 35E

The new equation is $x^{\text{'}}{}^2-y^{\text{'}}{}^2=2$

Explanation of Solution

Given:

  $xy=1$

Calculation:

For the product term to vanish, the angle of rotation of axes is given as

  $\alpha ={45}^{\circ }$

Now,

  $\begin{array}{l}\sin \alpha =\frac{\sqrt{2}}{2}\\ \cos \alpha =\frac{\sqrt{2}}{2}\end{array}$

After rotation of coordinate axes, the original coordinates, $x$ and $y$ , are replaced by new coordinates $x^{\text{'}}$ and $y^{\text{'}}$ , such that,

  $\begin{array}{l}x=x^{\text{'}}\cos \alpha -y^{\text{'}}\sin \alpha \\ =\frac{\sqrt{2}}{2}{x}^{\text{'}}-\frac{\sqrt{2}}{2}{y}^{\text{'}}\\ y=x^{\text{'}}\sin \alpha +y^{\text{'}}\cos \alpha \\ =\frac{\sqrt{2}}{2}{x}^{\text{'}}+\frac{\sqrt{2}}{2}{y}^{\text{'}}\end{array}$

Substituting the above values of $x$ and $y$ in the given hyperbola equation,

  $\left(\frac{\sqrt{2}}{2}{x}^{\text{'}}-\frac{\sqrt{2}}{2}{y}^{\text{'}}\right)\left(\frac{\sqrt{2}}{2}{x}^{\text{'}}+\frac{\sqrt{2}}{2}{y}^{\text{'}}\right)=1$

Simplify,

  $x^{\text{'}}{}^2-y^{\text{'}}{}^2=2$

Conclusion:

Hence, the new equation is $x^{\text{'}}{}^2-y^{\text{'}}{}^2=2$ .

(b)

To determine

To find: The new equation .

Answer to Problem 35E

The new equation is ${\left(x^{\text{'}}\right)}^2-{\left(y^{\text{'}}\right)}^2=2a$

Explanation of Solution

Given:

  $xy=a$

Calculation:

For the product term to vanish, the angle of rotation of axes is given as ,

  $\alpha ={45}^{\circ }$

Now,

  $\begin{array}{l}\sin \alpha =\frac{\sqrt{2}}{2}\\ \cos \alpha =\frac{\sqrt{2}}{2}\end{array}$

After rotation of coordinate axes, the original coordinates, $x$ and $y$ , are replaced by new coordinates $x^{\text{'}}$ and $y^{\text{'}}$ , such that,

  $\begin{array}{l}x=x^{\text{'}}\cos \alpha -y^{\text{'}}\sin \alpha \\ =\frac{\sqrt{2}}{2}{x}^{\text{'}}-\frac{\sqrt{2}}{2}{y}^{\text{'}}\\ y=x^{\text{'}}\sin \alpha +y^{\text{'}}\cos \alpha \\ =\frac{\sqrt{2}}{2}{x}^{\text{'}}+\frac{\sqrt{2}}{2}{y}^{\text{'}}\end{array}$

Substituting the above values of $x$ and $y$ in the given hyperbola equation,

  $\left(\frac{\sqrt{2}}{2}{x}^{\text{'}}-\frac{\sqrt{2}}{2}{y}^{\text{'}}\right)\left(\frac{\sqrt{2}}{2}{x}^{\text{'}}+\frac{\sqrt{2}}{2}{y}^{\text{'}}\right)=a$

Simplify.

  ${\left(x^{\text{'}}\right)}^2-{\left(y^{\text{'}}\right)}^2=2a$

Conclusion:

Hence, the new equation is ${\left(x^{\text{'}}\right)}^2-{\left(y^{\text{'}}\right)}^2=2a$ .