Question

Mark the correct alternative in each of the following:
For the ellipse $12x^2+4y^2+24x-16y+24=0$

• A. centries(-1,2)
  
• B. lengths of the axes are $\sqrt{3}$ and 1
  
• C. eccentricity = $\sqrt{\frac{2}{3}}$
  
• D. all of these

Answer 1

The correct option is D all of these

$12x^2+4y^2+24x-16y+24=0$
$\Rightarrow 12(x^2+2x)+4(y^2-4y)=-24$
$\Rightarrow 12(x^2+2x+1)+4(y^2-4y+4)=-24+12+16$
$\Rightarrow 12(x+1{)}^2+4(y-2{)}^2=-4$
$\Rightarrow \frac{(x+1{)}^2}{3}+\frac{(y-2{)}^2}{1}=1$
So, the centre is (-1,2)
Here, $a=\sqrt{3}$ and b=1
The length of the axes are $\sqrt{3}$ and 1.
Now,
$e=\sqrt{1-\frac{b^2}{a^2}}$
$e=\sqrt{1-\frac{1}{3}}$
$\Rightarrow e=\sqrt{\frac{2}{3}}$