Discussion Forum



1)

 The eccentricity of the ellipse y2+4x2-12x+6y+14=0 is

 


A) $\frac{1}{\sqrt{2}}$

B) $\frac{1}{2}$

C) $\frac{\sqrt{3}}{2}$

D) $\frac{1}{\sqrt{3}}$

Answer:

Option C

Explanation:

 y2+4x2-12x+6y+14=0

 4x2-12x+y2+6y+14=0

 $4(x^{2}-3x+\frac{9}{4})+y^{2}+6y+9=-14+9+9$

  $4(x-\frac{3}{2})^{2}+(y+3)^{2}=4$

  $\frac{(x-3/2)^{2}}{1}+\frac{(y+3)^{2}}{4}=1$

Here, a2=1, b2=4

$\therefore$    $e=\sqrt{1-\frac{a^{2}}{b^{2}}}$

  $e=\sqrt{1-\frac{1}{4}}=\frac{\sqrt{3}}{2}$