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A helicopter flying along the path y=7+x...

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

`(sqrt(5))/(6)`

B

`(1)/(3) sqrt((7)/(3))`

C

`(1)/(2)`

D

`(1)/(6)sqrt((7)/(3))`

Text Solution

Verified by Experts

The correct Answer is:
D
`y= 7+x^(3//2)`
Let the point on curve be `P(x_(1), 7+x_(1)^(3//2))` and given point be `A((1)/(2),7)` for the nearest point normal at P passes through A, so slope of line AP = slope of normal at P.
`(x_(1)^(3//2))/(x_(1)-(1)/(2))= -(dx)/(dy)= -(2)/(3sqrt(x_(1)))`
`3x_(1)^(2)=1-2x_(1)`
`3x_(1)^(2)+2x_(1)-1=0`
`(x_(1)+1)(3x_(1)-1)=0 `
`x_(1)=(1)/(3), [x_(1)=-1 ("not possible as " x_(1) gt 0)]`
Point `P((1)/(3), 7+(1)/(3sqrt(3)))`
So,`AP= sqrt((1)/(36)+(1)/(27))=(1)/(6) sqrt((7)/(3))`
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