Chapter B, Problem 52E

(a)

To determine

To calculate: To find the foci and asymptotes of the hyperbola $x^2-y^2=1$ and sketch its graph

Answer to Problem 52E

Foci of hyperbola are $\left(\sqrt{2},0\right)$ and $\left(-\sqrt{2},0\right)$

Asymptotes of hyperbola are $y=\pm x$

Explanation of Solution

Given information: Equation of hyperbola is $x^2-y^2=1$

Formula Used:

Standard equation of hyperbola is

  $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ ,

where $c^2=a^2+b^2$

Foci of above hyperbola is $\left(\pm c,0\right)$

Asymptotes of $y=\pm \frac{b}{a}x$

Calculation:

Equation of ellipse is given as

  $x^2-y^2=1$

Rewrite the equation as follows:

  $\frac{x^2}{1^2}-\frac{y^2}{1^2}=1$

On comparing the equation with standard equation of hyperbola,

  $a=1$

  $b=1$

Value of $c$ is calculated as follows:

  $\begin{array}{l}{c}^2=a^2+b^2\\ c^2=1^2+1^2\\ c^2=1+1\\ c^2=2\\ c=\pm \sqrt{2}\end{array}$

Thus, foci of hyperbola are $\left(\sqrt{2},0\right)$ and $\left(-\sqrt{2},0\right)$

Now, asymptotes of hyperbola are calculated as

  $\begin{array}{l}y=\pm \frac{1}{1}x\\ y=\pm x\end{array}$

On sketching the hyperbola:

  

Conclusion:

Hence, foci of hyperbola are $\left(\sqrt{2},0\right)$ and $\left(-\sqrt{2},0\right)$ and asymptotes of hyperbola are $y=\pm x$

(b)

To determine

To calculate: To sketch the graphof the hyperbola $y^2-x^2=1$

Answer to Problem 52E

The graph of hyperbola is sketched

Explanation of Solution

Given information: Equation of hyperbola is $x^2-y^2=1$

Formula Used:

Standard equation of hyperbola is

  $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ ,

where $c^2=a^2+b^2$

Foci of above hyperbola is $\left(\pm c,0\right)$

Asymptotes of $y=\pm \frac{b}{a}x$

Calculation:

Equation of ellipse is given as

  $y^2-x^2=1$

Foci and asymptotes are found out by reversing the roles of $x$ and $y$

Thus, foci are $\left(0,\pm c\right)$ and asymptotes are $y=\pm \frac{b}{a}x$

Using the values of equation $x^2-y^2=1$

Thus, foci of hyperbola are $\left(0,\sqrt{2}\right)$ and $\left(0,-\sqrt{2}\right)$

Now, asymptotes of hyperbola are calculated as

  $\begin{array}{l}y=\pm \frac{1}{1}x\\ y=\pm x\end{array}$

On sketching the hyperbola:

  

Conclusion:

Hence, graph of hyperbola is sketched