Let  be the point on x² = 8y which is nearest to the point (2, 4).


Now d is maximum or minimum when D is maximum or minimum.                                                  
      


           

Let (x, y) be any point on y = x² + 2 at which jet is at a particular moment.
∴  jet is at (x, x² + 2)
Let d be the distance between the jet at (x, x² + 2) and solider at (3, 2).
∴  d² = (x – 3)² + [ (x² + 2) – 2]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) = 12 x² + 2
At x = 1, f ' ' (x) = 12 + 2 = 14 > 0 ⇒ f (x) has a local minimum at x = 1
But x = 1 is only extreme point
∴ f (x) is minimum at x = 1
∴  nearest distance = d at x = 1

The equation of curve is
       
                                      ...(1)
Let D be the distance of any point (x, y) on the curve  from the origin (0, 0).




          

Now, 
Rejecting 



               
When 

Let (x, y) be the point on  which is nearest to the point (0, 5).

Let d be the distance between (0, 5) and 

Now d is maximum or minimum when D is maximum or minimum.
                     

Again at 

Let (x, y) be the point on  which is nearest to the point (–1, 2).

                             
                      



At