Question

analyze the equation. that is, find the center, vertices, and foci of the ellipse, and graph it.
$$ \frac { x ^ { 2 } } { 16 } + \frac { y ^ { 2 } } { 4 } = 1 $$
what are the coordinates of the center?
(type an ordered pair.)
what are the coordinates of the vertices?
(type an ordered pair. type an exact answer for each coordinate, using radicals as needed. use a comma to separa
what are the coordinates of the foci?
(type an ordered pair. type an exact answer for each coordinate, using radicals as needed. use a comma to separa
choose the correct graph.

Explanation:

Step1: Identify the standard form of the ellipse equation

The standard form of an ellipse equation is \(\frac{(x - h)^2}{a^2}+\frac{(y - k)^2}{b^2}=1\) (for a horizontally - oriented ellipse, \(a>b\)) or \(\frac{(x - h)^2}{b^2}+\frac{(y - k)^2}{a^2}=1\) (for a vertically - oriented ellipse, \(a>b\)). Given the equation \(\frac{x^{2}}{16}+\frac{y^{2}}{4}=1\), we can rewrite it as \(\frac{(x - 0)^2}{4^2}+\frac{(y - 0)^2}{2^2}=1\), where \(h = 0,k = 0,a = 4,b = 2\).

Step2: Find the center

For the ellipse equation \(\frac{(x - h)^2}{a^2}+\frac{(y - k)^2}{b^2}=1\), the center is \((h,k)\). Substituting \(h = 0\) and \(k = 0\) into the formula, the center is \((0,0)\).

Step3: Find the vertices

For a horizontally - oriented ellipse (\(\frac{(x - h)^2}{a^2}+\frac{(y - k)^2}{b^2}=1\)), the vertices are \((h\pm a,k)\). Since \(h = 0,k = 0,a = 4\), the vertices are \((4,0)\) and \((-4,0)\).

Step4: Find the foci

The relationship for the foci of an ellipse is \(c^{2}=a^{2}-b^{2}\). Substitute \(a = 4\) and \(b = 2\) into the formula: \(c^{2}=16 - 4=12\), so \(c=\sqrt{12}=2\sqrt{3}\). For a horizontally - oriented ellipse, the foci are \((h\pm c,k)\). Substituting \(h = 0,k = 0,c = 2\sqrt{3}\), the foci are \((2\sqrt{3},0)\) and \((-2\sqrt{3},0)\).

Answer:

• Center: \((0,0)\)
• Vertices: \((4,0),(-4,0)\)
• Foci: \((2\sqrt{3},0),(-2\sqrt{3},0)\)