Question

find an equation in standard form for the ellipse graphed

Explanation:

Step1: Identify the center of the ellipse

The center of the ellipse is at the origin \((0,0)\) since it's symmetric about both the x - axis and y - axis, and the mid - point of the vertices is \((0,0)\).

Step2: Determine the major and minor axes

• The ellipse is vertical (since it is taller along the y - axis). The standard form of a vertical ellipse centered at the origin is \(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}} = 1\), where \(a\) is the semi - major axis and \(b\) is the semi - minor axis, and \(a>b\).
• From the graph, the ellipse intersects the y - axis at \((0,4)\) and \((0, - 4)\), so the length of the semi - major axis \(a = 4\) (distance from center \((0,0)\) to \((0,4)\) or \((0, - 4)\)).
• The ellipse intersects the x - axis at \((2,0)\) and \((-2,0)\), so the length of the semi - minor axis \(b = 2\) (distance from center \((0,0)\) to \((2,0)\) or \((-2,0)\)).

Step3: Write the standard form equation

Substitute \(a = 4\) and \(b = 2\) into the standard form of the vertical ellipse equation \(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1\).
We get \(\frac{x^{2}}{2^{2}}+\frac{y^{2}}{4^{2}}=1\), which simplifies to \(\frac{x^{2}}{4}+\frac{y^{2}}{16}=1\).

Answer:

\(\frac{x^{2}}{4}+\frac{y^{2}}{16}=1\)