Chapter 9.3, Problem 38E

(a)

To determine

To find: The standard equation of hyperbola $3y^2-5x^2+6y-60x-192=0$ .

Answer to Problem 38E

The standard form of the hyperbola is. $\frac{{\left(y+1\right)}^2}{{\left(\sqrt{5}\right)}^2}-\frac{{\left(x+6\right)}^2}{{\left(\sqrt{3}\right)}^2}=1$

Explanation of Solution

Given information:

The equation of the hyperbola is $3y^2-5x^2+6y-60x-192=0$ .

Calculation:

Write the equation of the hyperbola.

  $\begin{array}{c}3y^2-5x^2+6y-60x-192=0\\ 3\left[y^2+2\left(1\right)\left(1\right)y+1^2\right]-5x^2-60x-192-3=0\\ 3\left[{\left(y+1\right)}^2\right]-5\left[x^2+12x+36+3\right]=0\\ 3{\left(y+1\right)}^2-5{\left(x+6\right)}^2=15\end{array}$

Simplify further.

  $\frac{{\left(y+1\right)}^2}{5}-\frac{{\left(x+6\right)}^2}{3}=1$

Therefore, the standard form of the hyperbola is. $\frac{{\left(y+1\right)}^2}{{\left(\sqrt{5}\right)}^2}-\frac{{\left(x+6\right)}^2}{{\left(\sqrt{3}\right)}^2}=1$ .

(b)

To determine

To find: The center, vertices, foci, asymptotes of hyperbola $3y^2-5x^2+6y-60x-192=0$ .

Answer to Problem 38E

The center of the hyperbola is $\left(-6,-1\right)$ , the vertices of the hyperbola is $\left(-6,-1+\sqrt{5}\right)$ , $\left(-6,-1-\sqrt{5}\right)$ , the foci of the hyperbola is $\left(-6,-1+2\sqrt{2}\right)$ and $\left(-6,-1-2\sqrt{2}\right)$ , the asymptotes of the hyperbola is $y=\frac{x\sqrt{15}}{3}+2\sqrt{15}-1$ , $y=-\frac{x\sqrt{15}}{3}-2\sqrt{15}-1$ .

Explanation of Solution

Given information:

Given equation of hyperbola is $3y^2-5x^2+6y-60x-192=0$ .

Calculation:

Write the equation of the hyperbola.

  $\begin{array}{c}3y^2-5x^2+6y-60x-192=0\\ 3\left[y^2+2\left(1\right)\left(1\right)y+1^2\right]-5x^2-60x-192-3=0\\ 3\left[{\left(y+1\right)}^2\right]-5\left[x^2+12x+36+3\right]=0\\ 3{\left(y+1\right)}^2-5{\left(x+6\right)}^2=15\end{array}$

Simplify further to find the standard equation of the hyperbola

  $\frac{{\left(y+1\right)}^2}{5}-\frac{{\left(x+6\right)}^2}{3}=1$

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

  $\begin{array}{l}a=\sqrt{5}\\ b=\sqrt{3}\\ h=-6\\ k=-1\end{array}$

The center of the hyperbola is given as $\left(-6,-1\right)$ .

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

So the vertices are $\left(-6,-1+\sqrt{5}\right)$ , $\left(-6,-1-\sqrt{5}\right)$ ,

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

Calculate the value of $c$ .

  $\begin{array}{c}{c}^2={\left(\sqrt{5}\right)}^2+{\left(\sqrt{3}\right)}^2\\ c^2=5+3\\ c^2=8\\ c=\sqrt{8}\end{array}$

Hence the foci are. $\left(-6,-1+2\sqrt{2}\right)$ and $\left(-6,-1-2\sqrt{2}\right)$ .

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

Substitute $\sqrt{5}$ for $a$ and $\sqrt{3}$ for $b$

  $-6$ for $h$ and $-1$ for $k$ to find the asymptotes.

. $y=\frac{x\sqrt{15}}{3}+2\sqrt{15}-1$ , $y=-\frac{x\sqrt{15}}{3}-2\sqrt{15}-1$ .

Therefore The center of the hyperbola is $\left(-6,-1\right)$ , the vertices of the hyperbola is $\left(-6,-1+\sqrt{5}\right)$ , $\left(-6,-1-\sqrt{5}\right)$ , the foci of the hyperbola is $\left(-6,-1+2\sqrt{2}\right)$ and $\left(-6,-1-2\sqrt{2}\right)$ , the asymptotes of the hyperbola is $y=\frac{x\sqrt{15}}{3}+2\sqrt{15}-1$ , $y=-\frac{x\sqrt{15}}{3}-2\sqrt{15}-1$ .

(c)

To determine

To draw: The graph of the hyperbola $3y^2-5x^2+6y-60x-192=0$ .

Answer to Problem 38E

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

Explanation of Solution

Given information:

The equation of the hyperbola is $3y^2-5x^2+6y-60x-192=0$ .

Calculation:

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

  $\begin{array}{c}3y^2-5x^2+6y-60x-192=0\\ 3\left[y^2+2\left(1\right)\left(1\right)y+1^2\right]-5x^2-60x-192-3=0\\ 3\left[{\left(y+1\right)}^2\right]-5\left[x^2+12x+36+3\right]=0\\ 3{\left(y+1\right)}^2-5{\left(x+6\right)}^2=15\end{array}$

Simplify further.

  $\begin{array}{c}3{\left(y+1\right)}^2=15+5{\left(x+6\right)}^2\\ {\left(y+1\right)}^2=\frac{15+5{\left(x+6\right)}^2}{3}\\ \left(y+1\right)=\pm \sqrt{\frac{15+5{\left(x+6\right)}^2}{3}}\\ y=\pm \sqrt{\frac{15+5{\left(x+6\right)}^2}{3}}-1\end{array}$

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 of the hyperbola using the graphical utility.

  

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

Hence, proved.