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Conic Sections — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsConic Sections1 Mark
MCQ
For the ellipse $12\text{x}^2+4\text{y}^2+24\text{x}-16\text{y}+25=0$
  • A
    centre is $(-1, 2)$
  • B
    lengths of the axes are $\sqrt{3}$ and $1$
  • C
    eccentricity $=\sqrt{\frac{2}{3}}$
  • ✓
    all of these.
✓

Answer

Correct option: D.
all of these.
$12\text{x}^2+4\text{y}^2+24\text{x}-16\text{y}+24=0$
$\Rightarrow12\big(\text{x}^2+2\text{x}\big)+4\big(\text{y}^2-4\text{y}\big)=-24$
$\Rightarrow12\big(\text{x}^2+2\text{x}+1\big)+4\big(\text{y}^2-4\text{y}+4\big)=-24+12+16$
$\Rightarrow12\big(\text{x}+1\big)^2+4\big(\text{y}-2\big)^2=4$
$\Rightarrow\frac{(\text{x}+1)^2}{3}+\frac{(\text{y}-2)^2}{1}=1$
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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