Question

what is the length of the minor axis of the ellipse \\( \frac{(x + 6)^{2}}{100}+\frac{y^{2}}{16}=1 \\)? write your answer in simplified, rationalized form.

Explanation:

Step1: Recall the standard form of an ellipse

The standard form of an ellipse is \(\frac{(x - h)^2}{a^2}+\frac{(y - k)^2}{b^2}=1\) (\(a>b\)) or \(\frac{(x - h)^2}{b^2}+\frac{(y - k)^2}{a^2}=1\) (\(a > b\)). The length of the minor axis is \(2b\).
For the given ellipse \(\frac{(x + 6)^2}{100}+\frac{y^2}{16}=1\), we have \(a^2 = 100\), \(b^2=16\).

Step2: Find the value of \(b\)

Since \(b^2 = 16\), then \(b=\sqrt{16}=4\).

Step3: Calculate the length of the minor axis

The length of the minor axis is \(2b\). Substituting \(b = 4\), we get \(2\times4=8\).

Answer:

\(8\)