Official solution
Given ellipse is \[{{{x^2}} \over 5} + {{{y^2}} \over 4} = 1\]
Let point P is \[(\sqrt 5 \cos \theta ,\,2\sin \theta )\]
\[{(PQ)^2}=5{\cos ^2}\theta + {(2\sin \theta + 4)^2}\]
\[ \Rightarrow \] (PQ)2 = \[5{\cos ^2}\theta + 4{\sin ^2}\theta + 16\sin \theta + 16\]
\[ \Rightarrow \] (PQ)2 = \[{\cos ^2}\theta + 4{\cos ^2}\theta + 4{\sin ^2}\theta + 16\sin \theta + 16\]
\[ \Rightarrow \] (PQ)2 = \[{\cos ^2}\theta + 4 + 16\sin \theta + 16\]
\[ \Rightarrow \] \[{(PQ)^2} = {\cos ^2}\theta + 16\sin \theta + 20\]
\[ \Rightarrow \] \[{(PQ)^2} = - {\sin ^2}\theta + 16\sin \theta + 21\]
= \[ - \left( {{{\sin }^2}\theta - 2.8\sin \theta + 64} \right) + 64 + 21\]
= \[85 - {(\sin \theta - 8)^2}\]
will be maximum when sin \[\theta \] = 1
\[ \Rightarrow {(PQ)^2}_{\max } = 85 - 49 = 36\]