Official solution
General point on \[{{{x^2}} \over 4} + {{{y^2}} \over 9} = 1\] is A(2cos\[\theta\], 3sin\[\theta\])
given B(-3, -5)
midpoint \[C\left( {{{2\cos \theta - 3} \over 2},{{3\sin \theta - 5} \over 2}} \right)\]
\[h = {{2\cos \theta - 3} \over 2};k = {{3\sin \theta - 5} \over 2}\]
\[ \Rightarrow {\left( {{{2h + 3} \over 2}} \right)^2} + {\left( {{{2k + 5} \over 3}} \right)^2} = 1\]
\[ \Rightarrow 36{x^2} + 16{y^2} + 108x + 80y + 145 = 0\]