Chapter 9, Problem 42CR

a.

To determine

To find: The standard form of the equation of hyperbola

Answer to Problem 42CR

The standard form of the equation of hyperbola is $\frac{{\left(y+3\right)}^2}{4}-\frac{{\left(x+1\right)}^2}{25}=1$

Explanation of Solution

Given information:

The given expression is $-4x^2+25y^2-8x+150y+121=0$

Calculation:

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

  $\begin{array}{l}-4x^2+25y^2-8x+150y+121=0\left[\text{Original equation}\right]\\ 25\left(y^2+6y\right)-4\left(x^2+2x\right)=-121\left[\begin{array}{l}\text{Group terms and factor 4 and 25 }\\ \text{out of }x\text{ and }y\text{ terms}\end{array}\right]\\ 25\left(y^2+6x+9\right)-4\left(x^2+2x+1\right)=-121++225-4\left[\text{Complete the squares}\right]\\ 25{\left(y+3\right)}^2-4{\left(x+1\right)}^2=100\left[\text{Write in completed square form}\right]\\ \frac{{\left(y+3\right)}^2}{4}-\frac{{\left(x+1\right)}^2}{25}=1\left[\text{Standard form}\right]\end{array}$

Therefore, the standard form of the equation of hyperbola is $\frac{{\left(y+3\right)}^2}{4}-\frac{{\left(x+1\right)}^2}{25}=1$

b.

To determine

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

Answer to Problem 42CR

The center, vertices, foci and asymptotes of the hyperbola are $\left(-1,-3\right)$ , $\left[\left(-1,-1\right),\left(-1,-5\right)\right]$ , $\left(-1,-3\pm \sqrt{29}\right)$ and $\frac{\sqrt{29}}{2}$

Explanation of Solution

Given information:

The given expression is $-4x^2+25y^2-8x+150y+121=0$

The standard form of the equation of hyperbola is $\frac{{\left(y+3\right)}^2}{4}-\frac{{\left(x+1\right)}^2}{25}=1$

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

  $a=2\text{ and }b=5$

The transverse axis is vertical where, $h=-1,k=-3,a=2,b=5$ and $c$ is obtained as,

  $\begin{array}{l}c=\sqrt{a^2+b^2}\\ c=\sqrt{2^2+5^2}\\ c=\sqrt{4+25}\\ c=\sqrt{29}\end{array}$

The center, vertices, foci and asymptote of the hyperbola can be obtained as,

The center of the hyperbola is $\left(h,k\right)=\left(-1,-3\right)$

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

The foci of the hyperbola is $\left(h,k\pm c\right)=\left(-1,-3\pm \sqrt{29}\right)$

The eccentricity $\left(e\right)=\frac{c}{a}=\frac{\sqrt{29}}{2}$

Therefore, the center, vertices, foci and asymptotes of the hyperbola are $\left(-1,-3\right)$ , $\left[\left(-1,-1\right),\left(-1,-5\right)\right]$ , $\left(-1,-3\pm \sqrt{29}\right)$ and $\frac{\sqrt{29}}{2}$

c.

To determine

To graph: The equation of hyperbola

Explanation of Solution

Given information:

The given expression is $-4x^2+25y^2-8x+150y+121=0$

Graph:

  

Interpretation: The equation is plotted and the graph is obtained