Question

The eccentricity of the conic $x^2-4x+4y^2=12$ is

• A. $\frac{\sqrt{3}}{2}$
• B. $\frac{2}{\sqrt{3}}$
• C. $\sqrt{3}$
• D. None of these

Answer 1

The correct option is A $\frac{\sqrt{3}}{2}$

Given, $x^2-4x+4y^2=12$

$\Rightarrow x^2-4x+4+4y^2=16$

$\Rightarrow {(x-2)}^2+4y^2=16$

$\Rightarrow \frac{{(x-2)}^2}{16}+\frac{y^2}{4}=1$

$\Rightarrow e^2=1-\frac{b^2}{a^2}=1-\frac{4}{16}=\frac{3}{4}$

$\Rightarrow e=\frac{\sqrt{3}}{2}$
So, option A is correct.