The equation of the ellipse having a vertex at $(6,1)$, a focus at $(4,1)$ and the eccentricity…
The equation of the ellipse having a vertex at $(6,1)$, a focus at $(4,1)$ and the eccentricity $\frac{3}{5}$ is
$\frac{(x-1)^2}{16}+\frac{(y-1)^2}{25}=1$
$\frac{(x-1)^2}{25}+\frac{(y-1)^2}{16}=1$
$\frac{(x+1)^2}{25}+\frac{(y+1)^2}{16}=1$
$\frac{(x+1)^2}{16}+\frac{(y+1)^2}{25}=1$
Solution
Let the equation of ellipse is,
$
\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1
$
Now,
$
\begin{array}{rlrl}
a-a e & =2 \\
\Rightarrow & a\left(1-\frac{3}{5}\right) & =2 \Rightarrow a=5 \\
& \text { So, } & b & =4
\end{array}
$
So,
Now, vertex comparing the vertex, we are getting
$
\text { and } \quad \begin{aligned}
& 6-h=5 \Rightarrow h=1 \\
& 1-k=0 \Rightarrow k=1
\end{aligned}
$
So, equation of required ellipse is
$
\frac{(x-1)^2}{25}+\frac{(y-1)^2}{16}=1 .
$