If $y = x$  and  $3y+2x=0$ are the equations 

of a pair of conjugate diameters of an ellipse, then the

  eccentricity of the ellipse is 

: Let the equation of the ellipse be  

$x^2/a^2+y^2/b^2=1$ 

Slope of the given diameters are  $m_1=1,m_2=-2/3$ 

$\Rightarrow m_1{m}_2=-2/3=-b^2/a^2$

[using the condition of conjugacy of two diameters] 

$\begin{array}{l}3b^2=2a^2\Rightarrow 3a^2\left(1-e^2\right)=2a^2\\ 1-e^2=2/3\Rightarrow e^2=1/3\Rightarrow e=1/\sqrt{3}\end{array}$