Chapter 9.3, Problem 37E

(a)

To determine

To find:The standard equation of hyperbola $2x^2-7y^2+16x+18=0$ .

Answer to Problem 37E

The standard form of the hyperbola is $\frac{{\left(x+4\right)}^2}{3^2}-\frac{y^2}{{\left(\sqrt{18/7}\right)}^2}=1$ .

Explanation of Solution

Given information:

The equation of the hyperbola is $2x^2-7y^2+16x+18=0$ .

Calculation:

Write the equation of the hyperbola.

  $\begin{array}{r}2x^2-7y^2+16x+18=0\\ 2\left(x^2+2\left(1\right)\left(4\right)x+16\right)-7y^2+18-72=0\\ 2{\left(x+4\right)}^2-7y^2=18\end{array}$

Simplify further.

  $\begin{array}{l}\frac{{\left(x+4\right)}^2}{9}-\frac{y^2}{\raisebox{1ex}{$18$}\!\left/ \!\raisebox{-1ex}{$7$}\right.}=1\\ \frac{{\left(x+4\right)}^2}{3^2}-\frac{y^2}{{\left(\sqrt{\raisebox{1ex}{$18$}\!\left/ \!\raisebox{-1ex}{$7$}\right.}\right)}^2}=1\end{array}$

Therefore, the standard form of the hyperbola is $\frac{{\left(x+4\right)}^2}{3^2}-\frac{y^2}{{\left(\sqrt{18/7}\right)}^2}=1$ .

(b)

To determine

To find:The center, vertices,foci, asymptotes of hyperbola $2x^2-7y^2+16x+18=0$ .

Answer to Problem 37E

The center of the hyperbola is $\left(-4,0\right)$ , the vertices of the hyperbola is $\left(-4+\sqrt{7},0\right)$ , $\left(-4-\sqrt{7},0\right)$ , the foci of the hyperbola is $\left(-1,0\right)$ and $\left(-7,0\right)$ , the asymptotes of the hyperbola is . $y=\frac{x\sqrt{14}}{7}+\frac{4\sqrt{14}}{7}$ ., $y=-\frac{x\sqrt{14}}{7}-\frac{4\sqrt{14}}{7}$ .

Explanation of Solution

Given information:

The equation of hyperbola is $2x^2-7y^2+16x+18=0$ .

Calculation:

Write the equation of the hyperbola.

  $\begin{array}{c}2x^2-7y^2+16x+18=0\\ 2\left(x^2+2\left(1\right)\left(4\right)x+16\right)-7y^2+18-72=0\\ 2{\left(x+4\right)}^2-7y^2=18\end{array}$

Simplify further.

  $\begin{array}{l}\frac{{\left(x+4\right)}^2}{9}-\frac{y^2}{\raisebox{1ex}{$18$}\!\left/ \!\raisebox{-1ex}{$7$}\right.}=1\\ \frac{{\left(x+4\right)}^2}{3^2}-\frac{y^2}{{\left(\sqrt{\raisebox{1ex}{$18$}\!\left/ \!\raisebox{-1ex}{$7$}\right.}\right)}^2}=1\end{array}$

Compare the given equation with the standard equation of hyperbola $\frac{{\left(x-h\right)}^2}{a^2}-\frac{{\left(y-k\right)}^2}{b^2}=1$ .

  $\begin{array}{l}a=3\\ b=\sqrt{\frac{18}{7}}\\ h=-4\\ k=0\end{array}$

The center of the hyperbola is given as $\left(-4,0\right)$ .

The vertices of the hyperbola is given by $\left(h+a,k\right)$ and $\left(h-a,k\right)$ .

So the vertices are $\left(-4+\sqrt{7},0\right)$ and $\left(-4-\sqrt{7},0\right)$ .

The foci of the hyperbola is given by $\left(h+c,k\right)$ and $\left(h-c,k\right)$ ,where $c^2=a^2+b^2$ . Calculate the value of $c$ .

  $\begin{array}{c}{c}^2=3^2+{\left(\sqrt{\frac{18}{7}}\right)}^2\\ c^2=9+\frac{18}{7}\\ c^2=\frac{34}{7}\\ c=\sqrt{\frac{34}{7}}\end{array}$

Hence the foci are $\left(-1,0\right)$ and $\left(-7,0\right)$ .

The asymptotes of the hyperbola is given by $y=\pm \frac{b}{a}\left(x-h\right)+k$ .

Substitute $3$ for $a$ and $\sqrt{\frac{18}{7}}$ for $b$

  $-4$ for $h$ and $0$ for $k$ to find the asymptotes.

  $y=\frac{x\sqrt{14}}{7}+\frac{4\sqrt{14}}{7}$ , $y=-\frac{x\sqrt{14}}{7}-\frac{4\sqrt{14}}{7}$ .

ThereforeThe center of the hyperbola is $\left(-4,0\right)$ , the vertices of the hyperbola is $\left(-4+\sqrt{7},0\right)$ , $\left(-4-\sqrt{7},0\right)$ , the foci of the hyperbola is $\left(-1,0\right)$ and $\left(-7,0\right)$ , the asymptotes of the hyperbola is $y=\frac{x\sqrt{14}}{7}+\frac{4\sqrt{14}}{7}$ , $y=-\frac{x\sqrt{14}}{7}-\frac{4\sqrt{14}}{7}$ .

(c)

To determine

To draw:The graph of the hyperbola $2x^2-7y^2+16x+18=0$ .

Answer to Problem 37E

The graph of the hyperbola is shown in figure (1).

Explanation of Solution

Given information:

The equation of the hyperbola is $2x^2-7y^2+16x+18=0$ .

Calculation:

To draw the graph of hyperbola, solve the equations of hyperbola in terms of $y$ .

  $\begin{array}{c}2x^2-7y^2+16x+18=0\\ 2\left(x^2+2\left(1\right)\left(4\right)x+16\right)-7y^2+18-72=0\\ 2{\left(x+4\right)}^2-7y^2=18\end{array}$

Simplify further.

  $\begin{array}{l}7y^2=2{\left(x+4\right)}^2-18\\ y^2=\frac{2{\left(x+4\right)}^2-18}{7}\end{array}$

Taking square root both the sides.

  $y=\pm \sqrt{\frac{2{\left(x+4\right)}^2-18}{7}}$

Use the above equation and the asymptotes to draw the graph of the hyperbola

Figure $\left(1\right)$

Therefore, the graph of the hyperbola is shown in figure (1).

Verify the result using the graphical utility:

Draw the graph for the hyperbola using the graphical utility.

From the above, it is clear that both the graphs are similar.

Hence, proved.