Question

The equation of the ellipse whose extremities of minor axis are $(3,1)$ and $(3,5)$ and eccentricity is $\frac{1}{2},$ is

• A. $4x^2+3y^2-18x-24y+47=0$
• B. $3x^2+4y^2-18x-24y+47=0$
• C. $3x^2+4y^2-24x-18y+47=0$
• D. $3x^2+4y^2+18x-24y-47=0$

Answer 1

The correct option is B $3x^2+4y^2-18x-24y+47=0$

As minor axis is parallel to $y-$axis
Hence required ellipse be as shown in the figure

$B^{\prime }=(3,1)$ and $B=(3,5)$, then $C=(3,3)$ will be the centre of the ellipse.
Let the equation of the ellipse be
$\frac{(x-3{)}^2}{a^2}+\frac{(y-3{)}^2}{b^2}=1(a>b)$
Now $B{B}^{\prime }=4$
$\therefore 2b=4\Rightarrow b^2=4$
as $a^2{e}^2=a^2-b^2$
$\Rightarrow \frac{a^2}{4}=a^2-b^2\Rightarrow a^2=\frac{16}{3}$
Hence equation of the ellipse will be $\frac{(x-3{)}^2}{\frac{16}{3}}+\frac{(y-3{)}^2}{4}=1$
$\Rightarrow 3x^2+4y^2-18x-24y+47=0$