Question

analyze this conic section to answer the questions below.
\frac{(x - 5)^2}{4}-\frac{y^2}{9}=1

c. ellipse: \frac{(x - h)^2}{a^2}+\frac{(y - k)^2}{b^2}=1 or \frac{(x - h)^2}{b^2}+\frac{(y - k)^2}{a^2}=1, a > b > 0
d. circle: \frac{(x - h)^2}{a^2}+\frac{(y - k)^2}{b^2}=1 or \frac{(x - h)^2}{b^2}+\frac{(y - k)^2}{a^2}=1, a = b
where is the center of this conic section? select the correct choice below and fill in any answer boxes in your choice.
a. (5,0) (type an ordered pair.)
b. the answer is undefined.
what are the values of a and b for this conic section? select the correct choice below and fill in any answer boxes in your choice.
a. a = \square, b = \square
b. the answer is undefined.

Explanation:

Step1: Identify the standard form of hyperbola

The given equation \(\frac{(x - 5)^{2}}{4}-\frac{y^{2}}{9}=1\) is in the form of a hyperbola \(\frac{(x - h)^{2}}{a^{2}}-\frac{(y - k)^{2}}{b^{2}}=1\), where \((h,k)\) is the center of the hyperbola.

Step2: Find the center \((h,k)\)

Comparing \(\frac{(x - 5)^{2}}{4}-\frac{y^{2}}{9}=1\) with \(\frac{(x - h)^{2}}{a^{2}}-\frac{(y - k)^{2}}{b^{2}}=1\), we have \(h = 5\) and \(k=0\). So the center is \((5,0)\).

Step3: Find the values of \(a\) and \(b\)

Since \(a^{2}=4\), then \(a=\sqrt{4}=2\) (we take the positive value as \(a>0\) in the standard form of the hyperbola). Since \(b^{2}=9\), then \(b = \sqrt{9}=3\) (we take the positive value as \(b>0\) in the standard form of the hyperbola).

Answer:

For the center of the conic - section: A. \((5,0)\)
For the values of \(a\) and \(b\): A. \(a = 2\), \(b=3\)