Question

suppose an ellipse with vertical and horizontal axes is centered at $(-3,1)$, and the distance from the center of the ellipse to a horizontal vertex is 8 and the distance from the center to a vertical vertex is 5. find an equation of the ellipse.

Explanation:

Step1: Recall the standard form of an ellipse equation

The standard form of an ellipse with center \((h,k)\) is \(\frac{(x - h)^2}{a^2}+\frac{(y - k)^2}{b^2}=1\) when the major axis is horizontal (\(a>b\)) or \(\frac{(x - h)^2}{b^2}+\frac{(y - k)^2}{a^2}=1\) when the major axis is vertical (\(a > b\)). Here, the center is \((h,k)=(-3,1)\). The distance from the center to a horizontal vertex is \(a = 8\) and the distance from the center to a vertical vertex is \(b = 5\).

Step2: Substitute the values into the standard - form equation

Substitute \(h=-3\), \(k = 1\), \(a = 8\), and \(b = 5\) into the standard - form equation \(\frac{(x - h)^2}{a^2}+\frac{(y - k)^2}{b^2}=1\).
We get \(\frac{(x+3)^2}{8^2}+\frac{(y - 1)^2}{5^2}=1\), which simplifies to \(\frac{(x + 3)^2}{64}+\frac{(y - 1)^2}{25}=1\).

Answer:

\(\frac{(x + 3)^2}{64}+\frac{(y - 1)^2}{25}=1\)