Chapter 9, Problem 41CR

a.

To determine

To find: The standard form of the equation of hyperbola

Answer to Problem 41CR

The standard form of the equation of hyperbola is $\frac{{\left(x-1\right)}^2}{16}-\frac{{\left(y+1\right)}^2}{9}=1$

Explanation of Solution

Given information:

The given expression is $9x^2-16y^2-18x-32y-151=0$

Calculation:

By completing the square, write the original equation in standard form

  $\begin{array}{l}9x^2-16y^2-18x-32y-151=0\left[\text{Original equation}\right]\\ 9\left(x^2-2x\right)-16\left(y^2+2y\right)=151\left[\begin{array}{l}\text{Group terms and factor 9 and 16 }\\ \text{out of }x\text{ and }y\text{ terms}\end{array}\right]\\ 9\left(x^2-2x+1\right)-16\left(y^2+2y+1\right)=151+9-16\left[\text{Complete the squares}\right]\\ 9{\left(x-1\right)}^2-16{\left(y+1\right)}^2=144\left[\text{Write in completed square form}\right]\\ \frac{{\left(x-1\right)}^2}{16}-\frac{{\left(y+1\right)}^2}{9}=1\left[\text{Standard form}\right]\end{array}$

Therefore, the standard form of the equation of hyperbola is $\frac{{\left(x-1\right)}^2}{16}-\frac{{\left(y+1\right)}^2}{9}=1$

b.

To determine

To find: The center, vertices, foci and asymptotes of the hyperbola

Answer to Problem 41CR

The center, vertices, foci and asymptotes of the hyperbola is $\left(1,-1\right)$ , $\left[\left(5,-1\right),\left(-3,-1\right)\right]$ , $\left[\left(6,-1\right),\left(-4,-1\right)\right]$ and $\frac{5}{4}$

Explanation of Solution

Given information:

The given expression is $9x^2-16y^2-18x-32y-151=0$

The standard form of the equation of hyperbola is $\frac{{\left(x-1\right)}^2}{16}-\frac{{\left(y+1\right)}^2}{9}=1$

Determine the value of $a\text{ and }b$ from the obtained standard equation of hyperbola

  $a=4\text{ and }b=3$

The transverse axis is horizontal where, $h=1,k=-1,a=4,b=3$ and $c$ is obtained as,

  $\begin{array}{l}c=\sqrt{a^2+b^2}\\ c=\sqrt{4^2+3^2}\\ c=\sqrt{16+9}\\ c=\sqrt{25}\\ c=5\end{array}$

The center of the hyperbola is $\left(h,k\right)=\left(1,-1\right)$

The vertices of the hyperbola is $\left(h,k\pm a\right)=\left(5,-1\right),\left(-3,-1\right)$

The foci of the hyperbola is $\left(h,k\pm c\right)=\left(6,-1\right),\left(-4,-1\right)$

The eccentricity $\left(e\right)=\frac{c}{a}=\frac{5}{4}$

Therefore, the center, vertices, foci and asymptotes of the hyperbola is $\left(1,-1\right)$ , $\left[\left(5,-1\right),\left(-3,-1\right)\right]$ , $\left[\left(6,-1\right),\left(-4,-1\right)\right]$ and $\frac{5}{4}$

c.

To determine

To graph: The equation of hyperbola

Explanation of Solution

Given information:

The given expression is $9x^2-16y^2-18x-32y-151=0$

Graph:

  

Interpretation: The equation is plotted and the graph is obtained