Question

question 8 of 10 which of the following is the equation of an ellipse centered at (5,1) having a vertical minor axis of length 4 and a major axis of length 6? a. $\frac{(x - 5)^2}{9}+\frac{(y - 1)^2}{4}=1$ b. $\frac{(x - 5)^2}{4}+\frac{(y - 1)^2}{9}=1$ c. $\frac{(x - 5)^2}{36}+\frac{(y - 1)^2}{16}=1$ d. $\frac{(x + 5)^2}{9}+\frac{(y + 1)^2}{4}=1$

Explanation:

Step1: Recall the standard - form of ellipse equation

The standard - form of an ellipse centered at \((h,k)\) is \(\frac{(x - h)^2}{a^2}+\frac{(y - k)^2}{b^2}=1\) for a horizontal major axis and \(\frac{(x - h)^2}{b^2}+\frac{(y - k)^2}{a^2}=1\) for a vertical major axis, where \((h,k)\) is the center of the ellipse, \(2a\) is the length of the major axis, and \(2b\) is the length of the minor axis.

Step2: Identify the values of \(h\), \(k\), \(a\), and \(b\)

The center of the ellipse is \((h,k)=(5,1)\). The length of the major axis \(2a = 6\), so \(a=3\). The length of the minor axis \(2b = 4\), so \(b = 2\). Since the minor axis is vertical, the major axis is horizontal. The equation of the ellipse is \(\frac{(x - 5)^2}{a^2}+\frac{(y - 1)^2}{b^2}=1\). Substituting \(a = 3\) and \(b = 2\) into the equation, we get \(\frac{(x - 5)^2}{9}+\frac{(y - 1)^2}{4}=1\).

Answer:

A. \(\frac{(x - 5)^2}{9}+\frac{(y - 1)^2}{4}=1\)