Question

analyze this conic section to answer the questions below.
\\( \frac{(x - 5)^{2}}{4}-\frac{y^{2}}{9}=1 \\)
what type of conic section is the equation?
a. parabola: \\( (y - k)^{2}=\pm 4 a(x - h) \\) or \\( (x - h)^{2}=\pm 4 a(y - k) \\)
b. hyperbola: \\( \frac{(x - h)^{2}}{a^{2}}-\frac{(y - k)^{2}}{b^{2}}=1 \\) or \\( \frac{(y - k)^{2}}{a^{2}}-\frac{(x - h)^{2}}{b^{2}}=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
a. (type an ordered pair.)
b. the answer is undefined.

Explanation:

Step1: Identify the conic section type

The general form of a hyperbola is \(\frac{(x - h)^{2}}{a^{2}}-\frac{(y - k)^{2}}{b^{2}} = 1\) (horizontal - transverse axis) or \(\frac{(y - k)^{2}}{a^{2}}-\frac{(x - h)^{2}}{b^{2}} = 1\) (vertical - transverse axis). The given equation \(\frac{(x - 5)^{2}}{4}-\frac{y^{2}}{9}=1\) is of the form \(\frac{(x - h)^{2}}{a^{2}}-\frac{(y - k)^{2}}{b^{2}} = 1\), so it is a hyperbola.

Step2: Find the center of the hyperbola

For a hyperbola of the form \(\frac{(x - h)^{2}}{a^{2}}-\frac{(y - k)^{2}}{b^{2}} = 1\), the center is \((h,k)\). In the equation \(\frac{(x - 5)^{2}}{4}-\frac{y^{2}}{9}=1\), comparing with \(\frac{(x - h)^{2}}{a^{2}}-\frac{(y - k)^{2}}{b^{2}} = 1\), we have \(h = 5\) and \(k=0\).

Answer:

1. B. Hyperbola: \(\frac{(x - h)^{2}}{a^{2}}-\frac{(y - k)^{2}}{b^{2}} = 1\) or \(\frac{(y - k)^{2}}{a^{2}}-\frac{(x - h)^{2}}{b^{2}} = 1\)
2. A. \((5,0)\)