Chapter 8.3, Problem 50E

Solution

To calculate: To calculate the vertices and foci , eccentricity and graph of an equation of the hyperbola.

The value of center is $\left(-3,1\right)$ .

The value of foci are $=\left(-3,1+\sqrt{34}\right),\left(-3,1-\sqrt{34}\right)$ .

The value of vertices are $\left(-3,4\right),\left(-3,-2\right)$ .

The value of eccentricity is $e=\frac{\sqrt{34}}{3}$ .

The graph of an equation is:

  

Given information: The given equation of a hyperbola is

  $25y^2-9x^2-50y-54x-281=0$ .

Calculation:

The given equation can be rewritten as in the form:

  $\begin{array}{c}25y^2-9x^2-50y-54x-281=0\\ 25y^2-50y-9x^2-54x=281\\ 25(y^2-2y+1)-9(x^2+6x+9)=281-81+25\\ 25{\left(y-1\right)}^2-9{\left(x+3\right)}^2=225\\ \frac{{\left(y-1\right)}^2}{9}-\frac{{\left(x+3\right)}^2}{25}=1\end{array}$

The standard form of the hyperbola is

  $\frac{{\left(y-1\right)}^2}{9}-\frac{{\left(x+3\right)}^2}{25}=1$

The general form of hyperbola is:

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

So, after comparing, we get,

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

The vertices in given general form can be written as:

Vertices $=\left(h,k-b\right),\left(h,k+b\right)$ .

Here, $\left(h,k\right)=\left(-3,1\right)$ .

The values of $\left(a,b\right)=\left(5,3\right)$ .

Then, the vertices will be calculated as:

  $\begin{array}{c}Vertices=\left(3,1-3\right),\left(3,1+3\right)\\ =\left(3,-2\right),\left(3,4\right)\end{array}$

The foci can be calculated as:

Foci $=\left(h,k-c\right),\left(h,k+c\right)$ .

The value of

  $c=\sqrt{a^2+b^2}$ .

  $\begin{array}{l}c=\sqrt{25+9}\\ c=\sqrt{34}\end{array}$

Thus, the value of foci is:

  $Foci=\left(-3,1+\sqrt{34}\right),\left(-3,1-\sqrt{34}\right)$

The eccentricity of a hyperbola is:

  $\begin{array}{l}e=\frac{c}{a}\\ e=\frac{\sqrt{34}}{3}\end{array}$

The graph of an equation is:

  

Thus, the values are :

Foci are $=\left(-3,1+\sqrt{34}\right),\left(-3,1-\sqrt{34}\right)$ .

Vertices are $\left(-3,4\right),\left(-3,-2\right)$ .

Center is $\left(-3,1\right)$ .

Eccentricity is $e=\frac{\sqrt{34}}{3}$ .