Chapter 9, Problem 52RE

a.

To determine

Find the standard form of the equation of the hyperbola.

Answer to Problem 52RE

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

Explanation of Solution

Given:

It is given in the question that the equation is $9x^2-y^2-72x+8y+119=0$ .

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 $9x^2-y^2-72x+8y+119=0$ .

Break the term in factor-ise form:

  $\begin{array}{l}9x^2-y^2-72x+8y+119=0\\ \Rightarrow (9x^2-72x)+(-y^2+8y)+119=0\\ \Rightarrow 9(x^2-8x)-(y^2-8y)=-119\\ \Rightarrow 9[(x^2-8x+16)-16]-[(y^2-8y+16)-16]=-119\\ \Rightarrow 9(x^2-8x+16)-144-(y^2-8y+16)+16=-119\\ $ \Rightarrow 9{(x-4)}^2-{(y-4)}^2=-119+144-16$$ \Rightarrow 9{(x-4)}^2-{(y-4)}^2=9$$ \Rightarrow {(x-4)}^2-\frac{{(y-4)}^2}{9}=1$\end{array}$

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

  $\begin{array}{l}$ \Rightarrow {(x-4)}^2-\frac{{(y-4)}^2}{9}=1$\Rightarrow \frac{{(x-4)}^2}{{(1)}^2}-\frac{{(y-4)}^2}{{(3)}^2}=1\end{array}$

Conclusion:

The standard form of equation is $\frac{{(x-4)}^2}{{(1)}^2}-\frac{{(y-4)}^2}{{(3)}^2}=1$ .

b.

To determine

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

Answer to Problem 52RE

The centre, vertices, foci and eccentricity are $(4,4);(5,4),(3,4);(4+\sqrt{10},4),(4-\sqrt{10},4);\sqrt{10}$ .

Explanation of Solution

Given: It is given in the question that the standard form is $\frac{{(x-4)}^2}{{(1)}^2}-\frac{{(y-4)}^2}{{(3)}^2}=1$ .

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

Now, the centre is $(4,4)$ .

Again, horizontal transverse axis , $a=1,b=3$ .

The vertices of the hyperbola is $(h\pm a,k)=(5,4),(3,4)$ .

Now, use a and b to find c.

  $c=\sqrt{a^2+b^2}=\sqrt{{(1)}^2+{(3)}^2}=\sqrt{1+9}=\sqrt{10}$

Then, foci of the hyperbola is $(h\pm c,k)=(4+\sqrt{10},4),(4-\sqrt{10},4)$ .

At last, the eccentricity of the hyperbola,

  $\begin{array}{l}e=\frac{c}{a}\\ e=\frac{\sqrt{10}}{\frac{1}{2}}=\sqrt{10}\end{array}$

Conclusion:

The centre, vertices, foci and eccentricity are $(4,4);(5,4),(3,4);(4+\sqrt{10},4),(4-\sqrt{10},4);\sqrt{10}$

c.

To determine

Draw the graph of the hyperbola.

Explanation of Solution

Given:

It is given in the question that the standard form is $\frac{{(x-4)}^2}{{(1)}^2}-\frac{{(y-4)}^2}{{(3)}^2}=1$ .

Graph:

  

Interpretation:

Here, the standard form of equation is $\frac{{(x-4)}^2}{{(1)}^2}-\frac{{(y-4)}^2}{{(3)}^2}=1$ , Put that equation in the graphing calculator , it will give the required graph in a proper manner.