Question

\\(\frac{x^2}{25} + \frac{y^2}{4} = 1\\)

the minor intercepts are at
\\(\circ\\) (\\(\pm5, 0\\))
\\(\circ\\) (\\(\pm2, 0\\))
\\(\circ\\) (\\(0, \pm2\\))

Explanation:

Step1: Recall ellipse standard form

The standard form of an ellipse centered at the origin is \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}} = 1\), where if \(a>b\), the major axis is along the \(x\)-axis and the minor axis is along the \(y\)-axis. The intercepts on the \(y\)-axis (minor intercepts when \(b < a\)) are found by setting \(x = 0\).

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

For the given ellipse \(\frac{x^{2}}{25}+\frac{y^{2}}{4}=1\), we have \(a^{2}=25\) (so \(a = 5\)) and \(b^{2}=4\) (so \(b = 2\)). Since \(a>b\), the minor axis is along the \(y\)-axis.

Step3: Find minor intercepts

To find the \(y\)-intercepts (minor intercepts here), set \(x = 0\) in the ellipse equation: \(\frac{0^{2}}{25}+\frac{y^{2}}{4}=1\), which simplifies to \(\frac{y^{2}}{4}=1\). Multiply both sides by 4: \(y^{2}=4\), then take square roots: \(y=\pm2\). So the minor intercepts are at \((0,\pm2)\).

Answer:

\((0, \pm 2)\)