Q.Find the equation of ellipse whose eccentricity is $\dfrac{2}{3}$, latus rectum is 5 and the centre is $(0, 0)$.

Standard ellipse (centre origin): $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$, with $b^2 = a^2(1 - e^2)$ and latus rectum $\dfrac{2b^2}{a}$.

From $e = \dfrac{2}{3}$: $b^2 = a^2\left(1 - \dfrac{4}{9}\right) = \dfrac{5a^2}{9}$.

From latus rectum $= 5$: $\dfrac{2b^2}{a} = 5 \Rightarrow b^2 = \dfrac{5a}{2}$.

Equating: $\dfrac{5a^2}{9} = \dfrac{5a}{2} \Rightarrow a = \dfrac{9}{2}$, so $a^2 = \dfrac{81}{4}$ and $b^2 = \dfrac{45}{4}$. …

For an ellipse centred at the origin with the major axis along the $x$-axis, the standard form is

where $a$ is the semi-major axis and $b$ the semi-minor axis. Two facts connect $a$ and $b$: the eccentricity satisfies $b^2 = a^2(1 - e^2)$, and the latus rectum has length $\frac{2b^2}{a}$.

1. Use the eccentricity. With $e = \frac{2}{3}$:

2. Use the latus rectum. Its length is $5$:

3. Equate (i) and (ii).

So $a^2 = \frac{81}{4}$.

4. Find $b^2$ from (ii):

5. Write the equation. …