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The equation of the ellipse having a vertex at (6, 1), a focus at (4, 1) and the eccentricity $\frac{3}{5}$ is

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$\frac{{(x - 1)}^{2}}{16} + \frac{{(y - 1)}^{2}}{25} = 1$

$\frac{(x - 1)^{2}}{25} + \frac{(y - 1)^{2}}{16} = 1$

$\frac{{(x + 1)}^{2}}{25} + \frac{{(y + 1)}^{2}}{16} = 1$

$\frac{{(x + 1)}^{2}}{16} + \frac{{(y + 1)}^{2}}{25} = 1$

answer is B.

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Detailed Solution

$A = (6, 1), S = (4, 1)$ major axis is parallel to x-axis

$e = \frac{3}{5}, A S = a - a e = 2$ $\Rightarrow a (1 - \frac{3}{5}) = 2 \Rightarrow a = 5$

$h + a = 6 \Rightarrow h = 1, b^{2} = a^{2} (1 - e^{2}) = 16 \Rightarrow k = 1$

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