Official solution
Centre at (0, 0)
\[{{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}}\] = 1
at point (4, - 1)
\[{{16} \over {{a^2}}} + {1 \over {{b^2}}}\] = 1
\[ \Rightarrow \] 16b2 + a2 = a2b2 . . . .(i)
at point (- 2, 2)
\[{4 \over {{a^2}}} + {4 \over {{b^2}}} = 1\]
\[ \Rightarrow \] 4b2 + 4a2 = a2b2 . . . .(ii)
\[ \Rightarrow \] 16b2 + a2 = 4a2 + 4b2
From equations (i) and (ii)
\[ \Rightarrow \] 3a2 = 12b2
\[ \Rightarrow \] a2 = 4b2
b2 = a2(1 - e2)
\[ \Rightarrow \] e2 = \[{3 \over 4}\]
\[ \Rightarrow \] e = \[{{\sqrt 3 } \over 2}\]