Question

graph the ellipse and locate the foci.
\\(\frac{x^2}{49} + \frac{y^2}{16} = 1\\)

choose the correct graph below.
\\(\bigcirc\\) a. graph of an ellipse \\(\bigcirc\\) b. graph of an ellipse \\(\bigcirc\\) c. graph of an ellipse

locate the foci.
(type ordered pairs. use a comma to separate answers. type exact answers, using radicals as needed. simplify your answers.)

Explanation:

Step1: Identify ellipse standard form

The given ellipse equation is \(\frac{x^2}{49}+\frac{y^2}{16} = 1\), which matches the standard form of an ellipse centered at the origin \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) where \(a > b>0\). Here, \(a^2 = 49\) so \(a = 7\), and \(b^2=16\) so \(b = 4\). Since \(a^2\) is under the \(x^2\) term, the major axis is along the \(x\)-axis.

Step2: Calculate the distance from center to foci (\(c\))

For an ellipse, the relationship between \(a\), \(b\), and \(c\) (distance from center to each focus) is \(c^2=a^2 - b^2\). Substitute \(a^2 = 49\) and \(b^2 = 16\):
\(c^2=49 - 16=33\), so \(c=\sqrt{33}\).

Step3: Determine the foci coordinates

Since the major axis is along the \(x\)-axis and the center is \((0,0)\), the foci are located at \((\pm c,0)\), which is \((\pm\sqrt{33},0)\).

Answer:

\((\sqrt{33},0),(-\sqrt{33},0)\)