If any point on an ellipse is (3sin\(\alpha\), 5cos\(\alpha\)), then what is the eccentricity of the ellipse?

Calculation:

Given any point on the ellipse is 3sinα, 5cosα. In the standard parametric form of an ellipse,

\(x = a\,\sinα,\quad y = b\,\cosα\)

we identify

\(a = 3,\quad b = 5 \)

Since (b > a), the semi-major axis is (b = 5) and the semi-minor axis is (a = 3). The eccentricity e of an ellipse is

\(e = \sqrt{\,1 - \frac{(\text{semi-minor})^{2}}{(\text{semi-major})^{2}}\,} = \sqrt{\,1 - \frac{a^{2}}{b^{2}}\,} \)

Substitute (a = 3) and (b = 5):

\(e = \sqrt{\,1 - \frac{3^{2}}{5^{2}}\,} = \sqrt{\,1 - \frac{9}{25}\,} = \sqrt{\frac{16}{25}} = \frac{4}{5} \)

Hence, the correct answer is Option 2.