Question

For the ellipse $x^2+3y^2+2x-12y+10=0,$ which of the following option(s) is/are correct

• A. centre is $(-1,2)$
• B. length of the axes are $2\sqrt{3}$ and $2$ units
• C. eccentricity is $\sqrt{\frac{2}{3}}$
• D. length of latus ractum is $\frac{2}{\sqrt{3}}$ units

Answer 1

Answer: (A), (B), (C), (D)

length of latus ractum is $\frac{2}{\sqrt{3}}$ units

$x^2+3y^2+2x-12y+10=0$
$\Rightarrow (x^2+2x)+3(y^2-4y)=-10$
$\Rightarrow (x^2+2x+1)+3(y^2-4y+4)=-10+1+12$
$\Rightarrow (x+1{)}^2+3(y-2{)}^2=3$
$\Rightarrow \frac{(x+1{)}^2}{3}+\frac{(y-2{)}^2}{1}=1$
So, the centre is $(-1,2)$.

Here, $a=\sqrt{3}$ and $b=1$
The length of the axis are $2\sqrt{3}$ and $2$ units.

And $e=\sqrt{1-\frac{b^2}{a^2}}$
$\Rightarrow e=\sqrt{1-\frac{1}{3}}$
$\Rightarrow e=\sqrt{\frac{2}{3}}$

Also length of latus ractum
$=\frac{2b^2}{a}=\frac{2}{\sqrt{3}}$ units