Question

the graph shows an ellipse. write its equation in standard form.

Explanation:

Step1: Find the center of the ellipse

The center of the ellipse is at the point \((h, k)\). From the graph, the center is at \((2, 7)\) (since it's 2 units to the right of the y - axis and 7 units up the x - axis).

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

• The standard form of an ellipse centered at \((h,k)\) is \(\frac{(x - h)^2}{a^2}+\frac{(y - k)^2}{b^2}=1\) (if it is a horizontal ellipse, where \(a\) is the semi - major axis and \(b\) is the semi - minor axis) or \(\frac{(y - k)^2}{a^2}+\frac{(x - h)^2}{b^2}=1\) (if it is a vertical ellipse).
• To find the length of the semi - major axis (\(a\)) and semi - minor axis (\(b\)):
• Let's count the number of grid units. The ellipse extends from \(x = - 2\) to \(x=6\) (centered at \(x = 2\)). The distance from the center to the rightmost point (semi - major axis length \(a\)) is \(6 - 2=4\).
• The ellipse extends from \(y = 5\) to \(y = 9\) (centered at \(y = 7\)). The distance from the center to the topmost point (semi - minor axis length \(b\)) is \(9 - 7 = 2\). Since the ellipse is wider along the x - axis, it is a horizontal ellipse.

Step3: Write the standard form equation

Using the standard form of a horizontal ellipse \(\frac{(x - h)^2}{a^2}+\frac{(y - k)^2}{b^2}=1\), where \(h = 2\), \(k=7\), \(a = 4\) and \(b = 2\).
Substitute these values into the equation:
\(\frac{(x - 2)^2}{4^2}+\frac{(y - 7)^2}{2^2}=1\)
Simplify the equation:
\(\frac{(x - 2)^2}{16}+\frac{(y - 7)^2}{4}=1\)

Answer:

\(\frac{(x - 2)^2}{16}+\frac{(y - 7)^2}{4}=1\)