Question

how long is the minor axis for the ellipse shown below?\\(\frac{(x + 4)^2}{25} + \frac{(y - 1)^2}{16} = 1\\)\
\\(\circ\\) 8\
\\(\circ\\) 18\
\\(\circ\\) 9\
\\(\circ\\) 12

Explanation:

Step1: Recall the standard form of an ellipse

The standard form of an ellipse is \(\frac{(x - h)^2}{a^2}+\frac{(y - k)^2}{b^2}=1\) (where \(a>b\), major axis is parallel to \(x\) - axis) or \(\frac{(x - h)^2}{b^2}+\frac{(y - k)^2}{a^2}=1\) (where \(a>b\), major axis is parallel to \(y\) - axis). For the given ellipse \(\frac{(x + 4)^2}{25}+\frac{(y - 1)^2}{16}=1\), we have \(a^2 = 25\), \(b^2=16\).

Step2: Find the length of the minor axis

The length of the minor axis of an ellipse is \(2b\). Since \(b^2 = 16\), then \(b = 4\). So the length of the minor axis is \(2b=2\times4\).

Answer:

\(8\)