The minimum distance of a point on the curve `y=x^2-4` from origin ,

Options

A. `sqrt(5)/2`
B. `sqrt(19)/2`
C. `sqrt(15/2)`
D. `sqrt(19/2)`

Answer

C

Explanation

Let P `(t, t^2 -4) ` be a point on `y=x^2-4` and let its distance from the origin be s. Then
`s=sqrt(t^2+(t^2-4)2)=sqrt(t^4-7 t^2+ 16)`
Let `z=s^2 ` Then
`Z=t^4 -7 t^2 + 16 rArr dz/dt = 4t^3-14 t and (d^2z)/(dt^2) =12t^2-14`
For maximum or minimum z we must have
`(dz)/(dt)=0 rArr 4t^3-14 t and (d^2z)/(dt^2)=12 t^2 -14 ` For maximum or minimum z, we must have `dz/dt =0 rArr 4t^3-14t =0 rArr t =0 ,pm sqrt(7/2)`
Clearly ,`d^2z/dt^2=28 gt 0 at t = pm sqrt(7/2)` So z is minimum at =`pmsqrt(7/2)`
For the value of t , `s=sqrt(7/2+ (7/2 -4)^2)=sqrt(7/2+1/4)=sqrt(15/2)`