Question

If the ellipse $2x^2+3y^2-4x-12y+k=0$ is passing through the point $(1,2+\frac{1}{\sqrt{3}}),$ then its eccentricity is

• A. $\sqrt{\frac{1}{2}}$
• B. $\frac{\sqrt{2}}{3}$
• C. $\sqrt{\frac{2}{3}}$
• D. $\sqrt{\frac{1}{3}}$

Answer 1

The correct option is D $\sqrt{\frac{1}{3}}$

$2x^2+3y^2-4x-12y+k=0$
$\Rightarrow 2(x-1{)}^2+3(y-2{)}^2=14-k$
Above ellipse passes through $(1,2+\frac{1}{\sqrt{3}})$
$\Rightarrow 0+1=14-k\Rightarrow k=13$
$\therefore$ Equation of ellipse is $2(x-1{)}^2+3(y-2{)}^2=1$
$\Rightarrow \frac{(x-1{)}^2}{\left(\frac{1}{2}\right)}+\frac{(y-2{)}^2}{\left(\frac{1}{3}\right)}=1$
$\Rightarrow a^2=\frac{1}{2},b^2=\frac{1}{3}$
$\therefore e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{\frac{1}{3}}$