X axis is the major axis and origin is the centre of an ellipse. If the distance between its directrices is \(\frac{18}{\sqrt{5}}\) and the ratio between the distances from the centre of this ellipse to its focus and its corresponding directrices is \(5:9\), then the length of its latus rectum is
Show Hint
Always verify if the ratio provided is \(ae:a/e\) or \(a/e:ae\) before computing eccentricity.
Step 1: Note what is given.
For this ellipse the distance between the directrices is $\dfrac{18}{\sqrt5}$, and the ratio of (centre to focus) to (centre to directrix) is $5:9$. We want the latus rectum. Step 2: Turn the ratio into eccentricity.
Centre to focus is $ae$; centre to directrix is $a/e$. Their ratio is $\dfrac{ae}{a/e}=e^2$. So $e^2=\dfrac59$, giving $e=\dfrac{\sqrt5}{3}$. Step 3: Use the directrix distance.
The distance between directrices is $\dfrac{2a}{e}=\dfrac{18}{\sqrt5}$. Step 4: Solve for $a$.
\[ \frac{2a}{\sqrt5/3}=\frac{18}{\sqrt5}\implies \frac{6a}{\sqrt5}=\frac{18}{\sqrt5}\implies 6a=18\implies a=3. \] Step 5: Find $b^2$.
Using $b^2=a^2(1-e^2)$: $b^2=9\left(1-\tfrac59\right)=9\cdot\tfrac49=4$. Step 6: Compute the latus rectum.
The latus rectum is $\dfrac{2b^2}{a}=\dfrac{2(4)}{3}=\dfrac{8}{3}$.
\[ \boxed{\dfrac{8}{3}} \]