Question

A helicopter flying along the path `y=7+x^((3)/(2))`, A soldier standint at point `((1)/(2),7)` wants to hit the helicopter when it is closest from him, then minimum distance is equal to (a) `(1)/(6)(sqrt2)/(3)` (b) `(1)/(2)` (c) `(1)/(3)sqrt((2)/(3))` (d) `sqrt((5)/(2))`
A. `(1)/(3)sqrt((7)/(3))`
B. `(sqrt5)/(6)`
C. `(1)/(6)sqrt((7)/(3))`
D. `(1)/(2)`

Answer 1

Correct Answer - C
The helicopter is nearest to the soldier, if the tangent ot the path, `y = x^( 3//2) + 7, (x ge 0)` of helicopter at point `(x, y)` is perpendicular to the line joining `(x, y)` and the position of soldier `((1)/(2),7)`.

`because ` Slope of tangent at point `(x, y)` is
`" " (dy)/(dx) = (3)/(2)x^(1//2) = m_1` (let) `" "` ...(i)
and slope of line joining `(x, y) and ((1)/(2), 7)` is
`" " m_2 = (y-7)/(x- (1)/(2)) " " `...(ii)
Now, `" " m_1*m_2 =-1 `
`rArr (3)/(2) x^(1//2) ((y- 7)/( x - (1//2)))=-1" " ` [ from Eqs. (i) and (ii)]
`rArr " " (3)/(2) x ^(1//2) = (x ^(3//2))/(x- (1)/(2)) = -1 " " [ because y = x^(3//2) = 7]`
`rArr " " (3)/(2) x^(2) = -x + (1)/(2) `
`rArr 3x^(2) + 3x - x - 1 = 0`
`rArr 3x(x + 1) - 1( x + 1) = 0`
`rArr " "x = (1)/(3), -1`
`because " "x ge 0`
`therefore " " x = (1)/(3) `
and So, `" " y = ((1)/(3))^(3//2) + 7 " " [ because y = x^(3//2) + 7]`
Thus, the nearest distance
`= sqrt(((1)/(2) - (1)/(3)) ^(2) + (7 - ((1)/(3))^(3//2) - 7)^(2)) = sqrt(((1)/(6))^(2) + ((1)/(3))^(3))`
`= sqrt((1)/(36)+ (1)/( 27)) = sqrt((3+ 4)/( 108)) = sqrt((7)/(108)) = (1)/(6) sqrt((7)/(3))`