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(B) Given,vertex $V = (6, 1)$ and focus $S = (4, 1)$. Since the $y$-coordinates are the same,the major axis is horizontal. The distance between the ve

Vedclass · Accepted Answer

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

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(B) Given,vertex $V = (6, 1)$ and focus $S = (4, 1)$. Since the $y$-coordinates are the same,the major axis is horizontal. The distance between the vertex and the focus is $a - ae = |6 - 4| = 2$. Given $e = \frac{3}{5}$,we have $a(1 - e) = 2$ $\Rightarrow a(1 - \frac{3}{5}) = 2$ $\Rightarrow a(\frac{2}{5}) = 2$ $\Rightarrow a = 5$. The center $(h, k)$ lies on the line $y = 1$. The distance from the center to the vertex is $a = 5$. Since the vertex is at $(6, 1)$ and the focus is at $(4, 1)$ (to the left of the vertex),the center must be at $(6 - 5, 1) = (1, 1)$. Now,$b^2 = a^2(1 - e^2) = 25(1 - \frac{9}{25}) = 25(\frac{16}{25}) = 16$. Thus,the equation of the ellipse is $\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$,which is $\frac{(x-1)^2}{25} + \frac{(y-1)^2}{16} = 1$.

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

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