# Find the eccentricity of the ellipse 2x2 + 3y2 = 6.

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Find the eccentricity of the ellipse 2x2 + 3y2 = 6.
1. $\frac{1}{{\sqrt 3 }}$
2. $\frac{2}{{\sqrt 3 }}$
3. $\frac{1}{{2\sqrt 3 }}$
4. $\sqrt 3$
5. None of these

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Correct Answer - Option 1 : $\frac{1}{{\sqrt 3 }}$

Concept:

The standard equation of the ellipse is $\frac{{{{\rm{x}}^2}}}{{{{\rm{a}}^2}}} + {\rm{\;}}\frac{{{{\rm{y}}^2}}}{{{{\rm{b}}^2}}} = 1$.

• For any ellipse the eccentricity is always less than 1.
• For any ellipse the following relation holds for:${{\rm{b}}^2} = {{\rm{a}}^2}\left( {1 - {{\rm{e}}^2}} \right)$

Calculation:

The given equation of the ellipse is 2x2 + 3y2 = 6.

Divide throughout by 6 to get the equation in the standard form.

$\frac{{{{\rm{x}}^2}}}{3} + {\rm{\;}}\frac{{{{\rm{y}}^2}}}{2} = 1$

Therefore, a2 = 3 and b2 = 2.

Now, for any ellipse using the relation we can write:

$2 = 3\left( {1 - {{\rm{e}}^2}} \right)$

e2$\frac{1}{3}$

e = $\frac{1}{{\sqrt 3 }}$

Therefore, the eccentricity of the given ellipse is $\frac{1}{{\sqrt 3 }}$