Question

review for final examination
score: 38/50 answered: 38/50
question 39
graph of an ellipse centered at the origin, with y-intercepts at (0, 8) and (0, -8), and x-intercepts at (-2, 0) and (2, 0)
write an equation for the ellipse graphed in standard form.
question help: video 1 video 2

Explanation:

Step1: Determine the center of the ellipse

The center \((h,k)\) of the ellipse is at the mid - point. Since the ellipse is symmetric about the \(y\) - axis (\(x = 0\)) and symmetric about the \(x\) - axis (mid - point between \(y = 8\) and \(y=-8\)). The center \((h,k)=(0,0)\)

Step2: Find the values of \(a\) and \(b\)

The standard form of an ellipse centered at \((h,k)\) is \(\frac{(x - h)^{2}}{b^{2}}+\frac{(y - k)^{2}}{a^{2}}=1\) (for a vertical major axis).
The distance from the center to the top (or bottom) of the ellipse is \(a\). The center is at \((0,0)\) and the top of the ellipse is at \(y = 8\), so \(a=8\).
The distance from the center to the right (or left) of the ellipse is \(b\). Let's assume the ellipse passes through \(x = 2\) (by looking at the grid). So \(b = 2\)

Step3: Write the equation of the ellipse

Substitute \(h = 0,k = 0,a = 8,b = 2\) into the standard form \(\frac{(x - h)^{2}}{b^{2}}+\frac{(y - k)^{2}}{a^{2}}=1\)
We get \(\frac{x^{2}}{4}+\frac{y^{2}}{64}=1\)

Answer:

\(\frac{x^{2}}{4}+\frac{y^{2}}{64}=1\)