If e is the eccentricity of an ellipse whose conjugate diameters are y = x and 3y = –2x then 3e² =

$$\begin{gathered} For conjugate diametre m_1{m}_2=\frac{-b^2}{a^2} \\ where m_1 and m_2 are slopes of the respective diameters \\ \left(1\right)\left(\frac{-2}{3}\right)=\frac{-b^2}{a^2} \\ \Rightarrow b^2=\frac{2}{3}{a}^2 \\ \Rightarrow a^2(1-e^2)=\frac{2}{3}{a}^2 \\ \Rightarrow (1-e^2)=\frac{2}{3} \\ \Rightarrow 3-3e^2=2 \\ \Rightarrow 3e^2=1 \end{gathered}$$