The point on the curve `y=sqrt(x-1)`, where the tangent is perpendicular to the line 2x+y-5=0 is

Options

A. (2,-1)
B. (10,3)
C. (2,1)
D. (5,-2)

Answer

C

Explanation

Let slope of the curve `y=sqrt(x-1)` is `m_(1)` and slope of the line 2x+y-5=0 is `m_(2)`
`m_(1)=(dy)/(dx)=(d)/(dx)sqrt(x-1)`
and `m_(2)=(dy)/(dx)=(d)/(dx)(5-2x)`
`impliesm_(1)=(1)/(2sqrt(x-1)) and m_(2)=-2`
We know that, lines are perpendicular if and only if `m_(1)m_(2)=-1`
`therefore(1)/(2sqrt(x-1))*(-2)=-1`
`implies(1)/(sqrt(x-1))=1implies sqrt(x-1)=1`
On squaring both sides, we get
`x-1=1impliesx=2`
On substituting x=2 in `y=sqrt(x-1)`, we get
Hence, coordination of the point on the curve, `y=sqrt(x-1)`, where tangent is perpendicular to the line 2x+y=5 is (2,1).