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

Question

Question

Answer 1

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 \(\frac{1}{{\sqrt 3 }}\).