Let (x, y) be any point on y = x² + 7 at which helicopter in at a particular moment.
∴  helicopter is at (x, x² + 7 ).
Let d be the distance between the jet at (x, x² + 7 ) and soldier at (3, 7).
∴  d² = (x – 3)² + (x² + 7 – 7)² = (x – 3)² + (x²)²
∴  d² = x⁴ + x² – 6 x + 9
Let f (x) = d² = x⁴ + x² – 6 x + 9
f ' (x) = 4 x³ + 2 x – 6 = 2 (2 x³ + x – 3) = 2 ( x – 1) (2 x² + 2 x + 3)
f ' (x) = 0 ⇒ 2 (x – 1) (2 x² + 2 x + 3) = 0

⇒ x = 1 as we reject imaginary values of x.
f ' ' (x) = 12x² + 2
At x = 1, f ' ' (x) = 12 + 2 – 14 > 0 ⇒ f (x) lies a local minimum at x = 1
But x – 1 is only extreme point
∴ f (x) is minimum at x = 1