Question

what is the length of the minor axis of an ellipse with equation \\(\frac{(x - 5)^2}{64} + \frac{(y + 8)^2}{16} = 1\\)? \\(\bigcirc\\) 32 \\(\bigcirc\\) 8 \\(\bigcirc\\) 16 \\(\bigcirc\\) 4 \\(\bigcirc\\) 64

Explanation:

Step1: Recall ellipse standard form

The standard form of an ellipse centered at \((h,k)\) is \(\frac{(x - h)^2}{a^2}+\frac{(y - k)^2}{b^2}=1\) (if major axis is horizontal) or \(\frac{(x - h)^2}{b^2}+\frac{(y - k)^2}{a^2}=1\) (if major axis is vertical), where \(a>b>0\). The length of the major axis is \(2a\) and the length of the minor axis is \(2b\).

Step2: Identify \(a^2\) and \(b^2\)

For the given ellipse \(\frac{(x - 5)^2}{64}+\frac{(y + 8)^2}{16}=1\), we have \(a^2 = 64\) and \(b^2=16\). So, \(a=\sqrt{64} = 8\) and \(b=\sqrt{16}=4\).

Step3: Determine minor axis length

Since \(a > b\) (because \(8>4\)), the major axis is horizontal and the minor axis is vertical. The length of the minor axis is \(2b\). Substituting \(b = 4\), we get \(2b=2\times4 = 8\)? Wait, no, wait. Wait, \(b^2 = 16\), so \(b = 4\)? Wait, no, wait, \(\frac{(x - 5)^2}{64}+\frac{(y + 8)^2}{16}=1\), so the denominator under \(x\) is \(a^2=64\) (so \(a = 8\)) and denominator under \(y\) is \(b^2 = 16\) (so \(b = 4\))? Wait, no, the standard form is \(\frac{(x - h)^2}{a^2}+\frac{(y - k)^2}{b^2}=1\) where \(a\) is the semi - major axis and \(b\) is the semi - minor axis when \(a>b\). So the length of the minor axis is \(2b\). Wait, but \(b^2=16\), so \(b = 4\)? Then \(2b=8\)? But wait, no, wait the options have 8, 16, etc. Wait, maybe I mixed up. Wait, the standard form: if the equation is \(\frac{(x - h)^2}{A}+\frac{(y - k)^2}{B}=1\), where \(A\) and \(B\) are positive. If \(A>B\), then the major axis is along the \(x\) - axis, semi - major axis \(a=\sqrt{A}\), semi - minor axis \(b=\sqrt{B}\), length of major axis \(2a\), length of minor axis \(2b\). If \(B>A\), major axis along \(y\) - axis, semi - major axis \(a=\sqrt{B}\), semi - minor axis \(b=\sqrt{A}\), length of major axis \(2a\), length of minor axis \(2b\).

In our case, \(A = 64\), \(B = 16\), so \(A>B\). So semi - major axis \(a=\sqrt{64}=8\), semi - minor axis \(b=\sqrt{16}=4\). Then length of minor axis is \(2b=2\times4 = 8\)? But wait, the options have 8 as an option. Wait, but let's check again. Wait, maybe I made a mistake. Wait, the denominator under \(y\) is 16, so \(b^2 = 16\), so \(b = 4\), so minor axis length is \(2b=8\). So the answer should be 8.

Answer:

8