Find the equation of ellipse whose latus-rectum is $5$ and eccentricity is $\dfrac{2}{3}$.
Latus-rectum of ellipse $= \dfrac{2b^2}{a} = 5$ (given)
$\implies$ $b^2 = \dfrac{5 a}{2}$ $\;\;\; \cdots \; (1)$
Eccentricity $= e = \dfrac{2}{3}$ (given)
For an ellipse, $\;$ $b^2 = a^2 \left(1 - e^2\right)$ $\;\;\; \cdots \; (2)$
Substituting the values of $b^2$ and $e$ in equation $(2)$ gives,
$\dfrac{5a}{2} = a^2 \left[1 - \left(\dfrac{2}{3}\right)^2\right]$
i.e. $\;$ $\dfrac{5}{2} = \dfrac{5a}{9}$
i.e. $\;$ $a = \dfrac{9}{2}$ $\implies$ $a^2 = \dfrac{81}{4}$
Substituting the value of $a$ in equation $(1)$ gives,
$b^2 = 5 \times \dfrac{9}{2} \times \dfrac{1}{2} = \dfrac{45}{4}$
The equation of ellipse is $\;\;$ $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$
i.e. $\;$ $\dfrac{x^2}{81 / 4} + \dfrac{y^2}{45 / 4} = 1$
i.e. $\;$ $\dfrac{4 x^2}{81} + \dfrac{4 y^2}{45} = 1$