The confidence interval for an exponential distribution is said to be: $$\frac{2n\bar{x}}{\chi^2_{1-\alpha /2,2n}}<\frac{1}{\lambda}<\frac{2n\bar{x}}{\chi^2_{\alpha /2,2n}}$$ In general we aim to obtain the shortest confidence interval possible. How can we be sure that this interval is the shortest? The presence of $\chi^2_{1-\alpha /2,2n}$ and $\chi^2_{\alpha /2,2n}$ suggests that there is a kind of a symmetry in the confidence interval, however, the exponential distribution is not symmetric.
Is not it true that we should build the interval around the region where the likelihood function is the greatest? It would mean that if the CI is $[a,b]$, then $L(a)=L(b)$ (likelihood) as we would be trying to utilise the part of likelihood function with the highest values.
Therefore, my question is - Is the CI presented at the beginning the shortest CI possible for exponential?