Find the equation of the ellipse if the center is $\left(3, -4\right)$, one of the foci is $\left(3 + \sqrt{3}, - 4\right)$ and $e = \dfrac{\sqrt{3}}{2}$
Let the center of the ellipse be $= C \left(h, k\right) = \left(3, -4\right)$
Let one of the foci be $= F_1 = \left(3 + \sqrt{3}, -4\right)$
Let the other focus be $= F_2 = \left(x_2, y_2\right)$
Now, center of the ellipse is the midpoint of $F_1 F_2$.
$\therefore$ $\;$ We have $\;$ $3 = \dfrac{3 + \sqrt{3} + x_2}{2}$; $\;$ $-4 = \dfrac{-4 + y_2}{2}$
i.e. $\;$ $x_2 = 3 - \sqrt{3}$; $\;$ $y_2 = -4$
$\therefore$ $\;$ $F_2 = \left(x_2, y_2\right) = \left(3 - \sqrt{3}, -4\right)$
$\therefore$ $\;$ From the given data, the major axis is parallel to the X axis.
$\therefore$ $\;$ Let the equation of the required ellipse be: $\;$ $\dfrac{\left(x - h\right)^2}{a^2} + \dfrac{\left(y - k\right)^2}{b^2} = 1$
Given: $\;$ $e = \dfrac{\sqrt{3}}{2}$
Now, $\;$ $F_1 F_2 = 2 a e$
i.e. $\;$ $\sqrt{\left(3 + \sqrt{3} - 3 + \sqrt{3}\right)^2 + \left(-4 + 4\right)^2} = 2 \times a \times \dfrac{\sqrt{3}}{2}$
i.e. $\;$ $2 \sqrt{3} = a \sqrt{3}$ $\implies$ $a = 2$
Further, $\;$ $b^2 = a^2 \left(1 - e^2\right)$
i.e. $\;$ $b^2 = \left(2\right)^2 \times \left[1 - \left(\dfrac{\sqrt{3}}{2}\right)^2\right]$
i.e. $\;$ $b^2 = 4 \times \dfrac{1}{4}$ $\implies$ $b^2 = 1$
$\therefore$ $\;$ The equation of the required ellipse is
$\dfrac{\left(x - 3\right)^2}{2^2} + \dfrac{\left(y + 4\right)^2}{2} = 1$
i.e. $\;$ $\dfrac{\left(x - 3\right)^2}{4} + \dfrac{\left(y + 4\right)^2}{1} = 1$