Question

analyze this conic section to answer the questions below
\\( \frac{(x - 5)^{2}}{4}-\frac{y^{2}}{9}=1 \\)
\\( \bigcirc \\) b. this conic section has a vertex at \\( (3,2) \\).
\\( \bigcirc \\) c. this conic section has a vertex at \\( (2,3) \\).
\\( \bigcirc \\) d. this conic section has vertices at \\( (7,0) \\) and \\( (3,0) \\).
what is the value of \\( c \\) for this conic section? select the correct choice below and fill in any answer boxes in your
choice.
\\( \bigcirc \\) a. \\( c=\sqrt{13} \\) (simplify your answer. type an exact answer, using radicals as needed.)
\\( \bigcirc \\) b. the answer is undefined
determine the focus or foci for the conic section.
\\( \bigcirc \\) a. the focus for this conic section is at \\( (5-\sqrt{13}, 5+\sqrt{13}) \\).
\\( \bigcirc \\) b. the foci for this conic section are at \\( (5-\sqrt{13}, 0) \\) and \\( (5+\sqrt{13}, 0) \\).
\\( \bigcirc \\) c. the foci for this conic section are at \\( (2,0) \\) and \\( (13,0) \\).
\\( \bigcirc \\) d. the focus for this conic section is at \\( (2,3) \\).

Explanation:

Step1: Recall the standard form of a hyperbola

The standard form of a hyperbola is \(\frac{(x - h)^2}{a^2}-\frac{(y - k)^2}{b^2}=1\), where \((h,k)\) is the center, \(a\) is the distance from the center to the vertices, and \(c^2=a^2 + b^2\) for hyperbola.
For the given equation \(\frac{(x - 5)^2}{4}-\frac{y^2}{9}=1\), we have \(h = 5,k = 0,a^2=4\), so \(a = 2\), \(b^2=9\), so \(b = 3\).

Step2: Calculate \(c\)

Using the formula \(c^2=a^2 + b^2\), substitute \(a = 2\) and \(b = 3\) into it.
\(c^2=4 + 9=13\), then \(c=\sqrt{13}\).

Step3: Find the foci

The foci of the hyperbola \(\frac{(x - h)^2}{a^2}-\frac{(y - k)^2}{b^2}=1\) are \((h\pm c,k)\).
Substitute \(h = 5,k = 0,c=\sqrt{13}\) into the formula.
The foci are \((5-\sqrt{13},0)\) and \((5+\sqrt{13},0)\).

Answer:

B. The foci for this conic section are at \((5-\sqrt{13},0)\) and \((5+\sqrt{13},0)\)