Chapter 9, Problem 48RE

a.

To determine

Find the standard form of the equation of the hyperbola.

Answer to Problem 48RE

The standard form of equation of the hyperbola is $\frac{x^2}{\frac{9}{4}}-\frac{y^2}{\frac{9}{4}}=1$ .

Explanation of Solution

Given:

It is given in the question that the equation is $x^2-y^2=\frac{9}{4}$ .

Concept Used:

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

  $x^2-y^2=\frac{9}{4}$

Calculation:

Here, the equation is.

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

  $\begin{array}{l}{x}^2-y^2=\frac{9}{4}\\ \frac{x^2}{\frac{9}{4}}-\frac{y^2}{\frac{9}{4}}=1\end{array}$

Conclusion:

The standard form of equation is $\frac{x^2}{\frac{9}{4}}-\frac{y^2}{\frac{9}{4}}=1$ .

b.

To determine

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

Answer to Problem 48RE

The centre, vertices, foci and eccentricity are $(0,0),(\pm \frac{3}{2},0),(\pm \frac{3\sqrt{2}}{2},0)and\sqrt{2}.$

Explanation of Solution

Given: It is given in the question that the standard form is $\frac{y^2}{4}-\frac{x^2}{5}=1$ .

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 $\frac{x^2}{\frac{9}{4}}-\frac{y^2}{\frac{9}{4}}=1\Rightarrow \frac{x^2}{{(\frac{3}{2})}^2}-\frac{y^2}{{(\frac{3}{2})}^2}=1$ .

Now, the centre is $(0,0)$ .

Again, horizontal transverse axis , $a=\frac{3}{2},b=\frac{3}{2}$ .

The vertices of the hyperbola is $(h\pm a,k)=(\pm \frac{3}{2},0)$ .

Now, use a and b to find c.

  $c=\sqrt{a^2+b^2}=\sqrt{{(\frac{3}{2})}^2+{(\frac{3}{2})}^2}=\sqrt{\frac{9}{4}+\frac{9}{4}}=\frac{3}{\sqrt{2}}=\frac{3\sqrt{2}}{2}$

Then, foci of the hyperbola is $(h\pm c,k)=(\pm \frac{3\sqrt{2}}{2},0)$ .

At last, the eccentricity of the hyperbola,

  $\begin{array}{l}e=\frac{c}{a}\\ e=\frac{\frac{3\sqrt{2}}{2}}{\frac{3}{2}}=\sqrt{2}\end{array}$

Conclusion:

The centre, vertices, foci and eccentricity are $(0,0),(\pm \frac{3}{2},0),(\pm \frac{3\sqrt{2}}{2},0)and\sqrt{2}.$

c.

To determine

Draw the graph of the hyperbola.

Explanation of Solution

Given:

It is given in the question that the standard form is $\frac{x^2}{\frac{9}{4}}-\frac{y^2}{\frac{9}{4}}=1$ .

Graph:

  

Interpretation:

Here, the standard form of equation is $\frac{x^2}{\frac{9}{4}}-\frac{y^2}{\frac{9}{4}}=1$ , Put that equation in the graphing calculator , it will give the required graph in a proper manner.