The point of the curve `y^(2)=2(x-3)` at which the normal is parallel to the line `y-2x+1=0` si

Options

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

Answer

C

Explanation

Given `y^(2)=2(x-3)" "....(i)`
On differentiating w.r.t., we get
`2y(dy)/(dx)=2`
`rArr (dy)/(dx)=(1)/(y)`
Stope of the normal `=(-1)/((dy//dx))=-y`
Slope of the given line =2
`:. y=-2`
From Eq. (i) x=5
`:.` Required points is (5,-2)