Mathematics | MHT CET 2020 | Discussion Page

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

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

$\frac{1}{2}$

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

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

Option C

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

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

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

Discussion Forum