Question

what is the length of the minor axis of the ellipse $\frac{x^{2}}{65}+\frac{y^{2}}{4}=1$? write your answer in simplified, rationalized form.

Explanation:

Step1: Recall the standard form of an ellipse

The standard form of an ellipse is \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}} = 1\). When \(a^{2}>b^{2}\), the length of the minor - axis is \(2b\).
For the ellipse \(\frac{x^{2}}{65}+\frac{y^{2}}{4}=1\), we have \(a^{2}=65\) and \(b^{2} = 4\).

Step2: Calculate the length of the minor - axis

Since \(b^{2}=4\), then \(b = 2\).
The length of the minor - axis is \(2b\). Substituting \(b = 2\) into \(2b\), we get \(2\times2=4\).

Answer:

\(4\)