Question

The equation of the ellipse having its centre at the point $(2,-3)$, one focus at $(3,-3)$ and one vertex at $(4,-3)$ is

• A. $\frac{(x+3{)}^2}{4}+\frac{(y-2{)}^2}{3}=1$
• B. $\frac{(x+3{)}^2}{3}+\frac{(y-2{)}^2}{4}=1$
• C. $\frac{(x-2{)}^2}{3}+\frac{(y+3{)}^2}{4}=1$
• D. $\frac{(x-2{)}^2}{4}+\frac{(y+3{)}^2}{3}=1$

Answer 1

The correct option is D $\frac{(x-2{)}^2}{4}+\frac{(y+3{)}^2}{3}=1$

Let $C=(2,-3),S=(3,-3)$ and $A=(4,-3)$
We know that mid point of foci is centre
Hence other focus will be $S^{\prime }\equiv (1,-3)$
$\therefore a=CA=2$
$\because$ foci are on the line parallel to $x-$axis
$\therefore$ distance between foci $2ae=2$
$a^2{e}^2=1=a^2-b^2\Rightarrow b^2=3$

$\therefore$ Equation of ellipse with centre as $(2,-3)$ and $a^2=4,b^2=3$ is
$\frac{(x-2{)}^2}{4}+\frac{(y+3{)}^2}{3}=1$