Chapter 9.6, Problem 10OE

To determine

To calculate: To find the graph of equation

Explanation of Solution

Given information: Equation is $\frac{{\left(x+1\right)}^2}{1}-\frac{{\left(y-2\right)}^2}{4}=1$

Formula Used:

The hyperbola with center $\left(0,0\right)$ and foci $\left(-c,0\right)$ and $\left(c,0\right)$ and difference of focal radii $2a$ has the equation:

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

Equation of the asymptotes are $y=\frac{b}{a}x$ and $y=-\frac{b}{a}x$

The hyperbola with center $\left(0,0\right)$ and foci $\left(0,-c\right)$ and $\left(0,c\right)$ and difference of focal radii $2a$ has the equation:

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

Equation of the asymptotes are $y=\frac{a}{b}x$ and $y=-\frac{a}{b}x$

Calculation:

Equation of hyperbola is given as follows:

  $\frac{{\left(x+1\right)}^2}{1}-\frac{{\left(y-2\right)}^2}{4}=1$

Rewriting the equation:

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

Comparing the equation with standard form of the equation of hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ :

  $\begin{array}{l}a=1\\ b=2\\ \left(h,k\right)=\left(-1,2\right)\end{array}$

Foci is calculated as follows:

  $\begin{array}{l}{b}^2=c^2-a^2\\ c^2=a^2+b^2\\ c^2=1^2+2^2\\ c^2=1+4\\ c^2=5\\ c=\pm \sqrt{5}\end{array}$

Thus, Foci are $\left(\sqrt{5},0\right)$ and $\left(-\sqrt{5},0\right)$

Also, asymptotes are calculated as follows:

  $y=\frac{b}{a}\left(x-h\right)+k$   and  $y=-\frac{b}{a}\left(x-h\right)+k$

Substituting the values:

  $y=2\left(x+1\right)+2$   and  $y=-2\left(x+1\right)+2$

Graph the hyperbola as follows:

  

Conclusion:

Hence, graph of hyperbola is