Question

find the standard form of the equation of the ellipse and give the location of its foci.
the standard form of the equation of the ellipse is \\(\frac{x^2}{9} + \frac{y^2}{16} = 1\\).
the foci are located at \\(square\\)
(type ordered pairs. use a comma to separate answers as needed. simplify your answers. type an exact answer, using radicals as needed.)

Explanation:

Step1: Recall ellipse focus formula

For an ellipse \(\frac{x^2}{b^2}+\frac{y^2}{a^2}=1\) (\(a > b>0\)), the distance from center to foci is \(c=\sqrt{a^2 - b^2}\), and foci are at \((0,\pm c)\) (since major axis is vertical).
Given equation \(\frac{x^2}{9}+\frac{y^2}{16}=1\), so \(a^2 = 16\), \(b^2=9\).

Step2: Calculate \(c\)

\(c=\sqrt{a^2 - b^2}=\sqrt{16 - 9}=\sqrt{7}\).

Step3: Determine foci coordinates

Center of ellipse is \((0,0)\) (since equation is in standard form with center at origin). Foci are on y - axis, so coordinates are \((0,\sqrt{7})\) and \((0,-\sqrt{7})\).

Answer:

The foci are located at \((0, \sqrt{7})\), \((0, -\sqrt{7})\) and the standard form of the ellipse equation is \(\frac{x^2}{9}+\frac{y^2}{16}=1\).