Chapter 9, Problem 51RE

a.

To determine

Find the standard form of the equation of the hyperbola.

Answer to Problem 51RE

The standard form of equation of the hyperbola is $\frac{{(x+6)}^2}{(\frac{101}{2})}-\frac{{(y-1)}^2}{202}=1$ .

Explanation of Solution

Given:

It is given in the question that the equation is $y^2-4x^2-2y-48x+59=0$ .

Concept Used:

In this, use the concept of dividing the Right Hand term to Left hand term make the equation like this, $\frac{x^2}{{(a)}^2}-\frac{y^2}{{(b)}^2}=1$ .

Calculation:

Here, the equation is $y^2-4x^2-2y-48x+59=0$ .

Break the term in factor-ise form:

  $\begin{array}{l}{y}^2-4x^2-2y-48x+59=0\\ \Rightarrow y^2-2y-4x^2-48x+59=0\\ \Rightarrow y^2-2y-4(x^2+12x)=-59\\ \Rightarrow [(y^2-2y+1)-1]-[4(x^2+12x+36)-36]=-59\\ $ \Rightarrow {(y-1)}^2-1-4{(x+6)}^2-144=-59$$ \Rightarrow {(y-1)}^2-4{(x+6)}^2=-59+144+1$$ \Rightarrow {(y-1)}^2-4{(x+6)}^2=86$\end{array}$

Divide the RHS term to the LHS side and make RHS one.

  $\begin{array}{l}$ {(y-1)}^2-4{(x+6)}^2=86$\frac{{(x+6)}^2}{(\frac{101}{2})}-\frac{{(y-1)}^2}{202}=1\end{array}$

Conclusion:

The standard form of equation is $\frac{{(x+6)}^2}{(\frac{101}{2})}-\frac{{(y-1)}^2}{202}=1$ .

b.

To determine

Calculate the center, vertices, foci and eccentricity of the hyperbola.

Answer to Problem 51RE

The centre, vertices, foci and eccentricity are $(-6,1);(-6\pm \frac{\sqrt{202}}{2});(-6\pm \frac{\sqrt{1010}}{2});\sqrt{5}$ .

Explanation of Solution

Given: It is given in the question that the standard form is $\frac{{(x+6)}^2}{(\frac{101}{2})}-\frac{{(y-1)}^2}{202}=1$ .

Concept Used:

In this, use the concept of finding the centre, vertices by $(h,k\pm a)$ ,foci by $(h,k\pm c)$ and eccentricity by $e=\frac{c}{a}$ by understanding the standard form.

Calculation:

Here, the standard form is $\frac{{(x+6)}^2}{(\frac{101}{2})}-\frac{{(y-1)}^2}{202}=1\Rightarrow \frac{{(x+6)}^2}{\sqrt{\frac{101}{2}}}-\frac{{(y-1)}^2}{\sqrt{202}}=1$ .

Now, the centre is $(-6,1)$ .

Again, horizontal transverse axis , $a=\sqrt{\frac{101}{2}},b=\sqrt{202}$ .

The vertices of the hyperbola is $(h\pm a,k)=(-6\pm \frac{\sqrt{202}}{2})$ .

Now, use a and b to find c.

  $c=\sqrt{a^2+b^2}=\sqrt{{(\frac{\sqrt{101}}{2})}^2+{(\sqrt{202})}^2}=\frac{\sqrt{1010}}{2}$

Then, foci of the hyperbola is $(h\pm c,k)=(-6\pm \frac{\sqrt{1010}}{2})$ .

At last, the eccentricity of the hyperbola,

  $\begin{array}{l}e=\frac{c}{a}\\ e=\frac{\frac{\sqrt{1010}}{2}}{\frac{\sqrt{101}}{2}}=\sqrt{5}\end{array}$

Conclusion:

The centre, vertices, foci and eccentricity are $(-6,1);(-6\pm \frac{\sqrt{202}}{2});(-6\pm \frac{\sqrt{1010}}{2});\sqrt{5}$

c.

To determine

Draw the graph of the hyperbola.

Explanation of Solution

Given:

It is given in the question that the standard form is $\frac{{(x+6)}^2}{(\frac{101}{2})}-\frac{{(y-1)}^2}{202}=1$ .

Graph:

  

Interpretation:

Here, the standard form of equation is $\frac{{(x+6)}^2}{(\frac{101}{2})}-\frac{{(y-1)}^2}{202}=1$ , Put that equation in the graphing calculator , it will give the required graph in a proper manner.