Question

exam 3
100 points possible answered: 10/14
question 11
find an equation in standard form for the ellipse graphed below.
(graph of an ellipse on a coordinate grid with x-axis from -6 to 6 and y-axis from -6 to 6, centered at (0,0), vertical major axis with vertices at (0,2) and (0,-2), horizontal minor axis with co - vertices at (1,0) and (-1,0))
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Explanation:

Step1: Identify the center of the ellipse

The center of the ellipse is at the origin \((0,0)\) since it is symmetric about both the x - axis and y - axis and the center is at the intersection of the major and minor axes.

Step2: Determine the length of the semi - major and semi - minor axes

• For the x - axis: The ellipse extends from \(x=- 1\) to \(x = 1\), so the length of the semi - minor axis (along the x - axis) \(b=1\) (because the distance from the center \((0,0)\) to the vertex on the x - axis is \(1\)).
• For the y - axis: The ellipse extends from \(y=-2\) to \(y = 2\), so the length of the semi - major axis (along the y - axis) \(a = 2\) (because the distance from the center \((0,0)\) to the vertex on the y - axis is \(2\)).

Step3: Recall the standard form of the equation of an ellipse

The standard form of the equation of an ellipse with center \((h,k)=(0,0)\), major axis along the y - axis is \(\frac{(x - h)^{2}}{b^{2}}+\frac{(y - k)^{2}}{a^{2}}=1\), where \(a>b\).

Substituting \(h = 0\), \(k = 0\), \(a = 2\) and \(b = 1\) into the formula, we get:

\(\frac{x^{2}}{1^{2}}+\frac{y^{2}}{2^{2}}=1\), which simplifies to \(\frac{x^{2}}{1}+\frac{y^{2}}{4}=1\) or \(x^{2}+\frac{y^{2}}{4}=1\)

Answer:

\(x^{2}+\frac{y^{2}}{4}=1\)