Chapter 9, Problem 50RE

a.

To determine

Find the standard form of the equation of the hyperbola.

Answer to Problem 50RE

The standard form of equation of the hyperbola is $\frac{\text{(y+3)²}}{4}-\frac{\text{(x+1)²}}{25}=1$ .

Explanation of Solution

Given:

It is given in the question that the equation is $\text{-4x² + 25y² - 8x + 150y + 121 = 0}$ .

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 $\text{-4x² + 25y² - 8x + 150y + 121 = 0}$ .

Break the term in factor-ise form:

  $\begin{array}{l}\text{-4x² + 25y² - 8x + 150y + 121 = 0}\\ \text{=> (25y²}+150y)+(-4x^2-8x)+121=0\\ \text{=>25(y²+6y) -4(x²+2x) =-121}\\ \text{=>25[(y²+6y +9)-9] - 4[(x²+2x+1) -1)] =-121}\\ \text{=>25(y²+6y +9})-225-4(\text{x²+2x+1})+4=-121\\ $ \Rightarrow 25{(y+3)}^2-4{(x+1)}^2=-121+225-4$$ \Rightarrow 25{(y+3)}^2-4{(x+1)}^2=100$\end{array}$

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

  $\begin{array}{l}$ 25{(y+3)}^2-4{(x+1)}^2=100$\frac{\text{(y+3)²}}{4}-\frac{\text{(x+1)²}}{25}=1\end{array}$

Conclusion:

The standard form of equation is $\frac{\text{(y+3)²}}{4}-\frac{\text{(x+1)²}}{25}=1$ .

b.

To determine

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

Answer to Problem 50RE

The centre, vertices, foci and eccentricity are $(-1,-3);(-1,-1),(-1,5);(-1,-3+\sqrt{29}),(-1,-3-\sqrt{29});\frac{\sqrt{29}}{2}$

Explanation of Solution

Given: It is given in the question that the standard form is $\frac{\text{(y+3)²}}{4}-\frac{\text{(x+1)²}}{25}=1$ .

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 $\frac{\text{(y+3)²}}{4}-\frac{\text{(x+1)²}}{25}=1\Rightarrow \frac{\text{[y-(-3)]²}}{{(2)}^2}-\frac{\text{[x-(-1)]²}}{{(5)}^2}=1$ .

Now, the centre is $(-1,-3)$ .

Again, horizontal transverse axis , $a=2,b=5$ .

The vertices of the hyperbola is $(h,k\pm a)=(-1,-1),(-1,5)$ .

Now, use a and b to find c.

  $c=\sqrt{a^2+b^2}=\sqrt{{(2)}^2+{(5)}^2}=\sqrt{4+25}=\sqrt{29}$

Then, foci of the hyperbola is $(h,k\pm c)=(-1,-3+\sqrt{29}),(-1,-3-\sqrt{29})$ .

At last, the eccentricity of the hyperbola,

  $\begin{array}{l}e=\frac{c}{a}\\ e=\frac{\sqrt{29}}{2}\end{array}$

Conclusion:

The centre, vertices, foci and eccentricity are $(-1,-3);(-1,-1),(-1,5);(-1,-3+\sqrt{29}),(-1,-3-\sqrt{29});\frac{\sqrt{29}}{2}$

c.

To determine

Draw the graph of the hyperbola.

Explanation of Solution

Given:

It is given in the question that the standard form is $\frac{\text{(y+3)²}}{4}-\frac{\text{(x+1)²}}{25}=1$ .

Graph:

  

Interpretation:

Here, the standard form of equation is $\frac{\text{(y+3)²}}{4}-\frac{\text{(x+1)²}}{25}=1$ , Put that equation in the graphing calculator , it will give the required graph in a proper manner.