If $$y = x$$  and $$3y + 2x = 0$$   are the equations of a pair of conjugate diameters of the ellipse $$\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$$   then its eccentricity is :

Solution :
We know that two diameters $$y = {m_1}x,\,y = {m_2}x$$     are conjugate diameters of $$\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$$   if $${m_1}{m_2} = - \frac{{{b^2}}}{{{a^2}}}$$
Hence, we have $$1.\left( { - \frac{2}{3}} \right) = - \frac{{{b^2}}}{{{a^2}}}{\text{ or }}{b^2} = \frac{2}{3}{a^2}$$
$$\therefore \,\,{b^2} = {a^2}\left( {1 - {e^2}} \right)\,\, \Rightarrow \frac{2}{3}{a^2} = {a^2}\left( {1 - {e^2}} \right)\,\, \Rightarrow \frac{2}{3} = 1 - {e^2}$$