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The equation of ellipse if length of L.R. is 5 and eccentricity is 2/3 is

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$\begin{aligned} \frac{2 x^{2}}{81} + \frac{2 y^{2}}{45} = 1 \end{aligned}$

$\begin{aligned} \frac{4}{81} x^{2} + \frac{4}{45} y^{2} = 1 \end{aligned}$

$\frac{4}{27} x^{2} + \frac{4}{18} y^{2} = 1$

$\frac{2 x^{2}}{27} + \frac{2 y^{2}}{18} = 1$

answer is B.

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Detailed Solution

$\begin{array}{l} \frac{2 b^{2}}{a} = 5 \Rightarrow 2 b^{2} = 5 a \\ e^{2} = 1 - \frac{b^{2}}{a^{2}} ⇒ \frac{4}{9} = 1 - \frac{5 a}{2 a^{2}} \\ ∴ \frac{5}{9} = \frac{5}{2 a} \Rightarrow a = \frac{9}{2} \end{array}$

$∴ b^{2} = \frac{5 a}{2} = \frac{45}{4}$

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