MCQ
For the ellipse $12\text{x}^2+4\text{y}^2+24\text{x}-16\text{y}+25=0$
- Acentre is (-1, 2)
- Blengths of the axes are $\sqrt{3}$ and 1
- Ceccentricity $=\sqrt{\frac{2}{3}}$
- Dall of these.
Solution:
$12\text{x}^2+4\text{y}^2+24\text{x}-16\text{y}+24=0$
$\Rightarrow12\big(\text{x}^2+2\text{x}\big)+4\big(\text{y}^2-4\text{y}\big)=-24$
$\Rightarrow12\big(\text{x}^2+2\text{x}+1\big)+4\big(\text{y}^2-4\text{y}+4\big)=-24+12+16$
$\Rightarrow12\big(\text{x}+1\big)^2+4\big(\text{y}-2\big)^2=4$
$\Rightarrow\frac{(\text{x}+1)^2}{3}+\frac{(\text{y}-2)^2}{1}=1$
So, the centre is a
$(-1,\ 2).$Here,
$\text{a}=\sqrt{3}$ and $\text{b}=1$The lengths of the axes are
$\sqrt{3}$ and 1.Now,
$\text{e}=\sqrt{1-\frac{\text{b}62}{\text{a}^2}}$$\text{e}=\sqrt{1-\frac{1}{3}}$
$\Rightarrow\text{e}=\sqrt{\frac{2}{3}}$
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