Chapter 9, Problem 47RE

a.

To determine

Find the standard form of the equation of the hyperbola.

Answer to Problem 47RE

The standard form of equation of the hyperbola is $\frac{y^2}{4}-\frac{x^2}{5}=1$ .

Explanation of Solution

Given:

It is given in the question that the equation is $5y^2-4x^2=20$ .

Concept Used:

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

Calculation:

Here, the equation is $5y^2-4x^2=20$ .

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

  $\begin{array}{l}5y^2-4x^2=20\\ \frac{5y^2}{20}-\frac{4x^2}{20}=1\\ \frac{y^2}{4}-\frac{x^2}{5}=1\end{array}$

Conclusion:

The standard form of equation is $\frac{y^2}{4}-\frac{x^2}{5}=1$ .

b.

To determine

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

Answer to Problem 47RE

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

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 $\frac{y^2}{4}-\frac{x^2}{5}=1$

  $\Rightarrow \frac{y^2}{{(2)}^2}-\frac{x^2}{{(5)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}=1$ .

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

Again, vertical transverse axis , $a=2,b={(5)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$ .

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

Now, use a and b to find c.

  $c=\sqrt{a^2+b^2}=\sqrt{{(2)}^2+{(\sqrt{5})}^2}=\sqrt{4+5}=\sqrt{9}=3$

Then, foci of the hyperbola is $(h,k\pm c)=(0,\pm 3)$ .

At last, the eccentricity of the hyperbola,

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

Conclusion:

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

c.

To determine

Draw the graph of the hyperbola.

Explanation of Solution

Given:

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

Graph:

  

Interpretation:

Here, the standard form of equation is $\frac{y^2}{4}-\frac{x^2}{5}=1$ , Put that equation in the graphing calculator , it will give the required graph in a proper manner.