If the normal line at `(1, -2)` on the curve `y^2=5x-1` is `ax - 5y +b=0` then the values of `a and b` are

Options

A. 4,-14
B. 4,14
C. `-4,14`
D. `4,2`

Answer

A

Explanation

Given `y^(2)=5x-1` ……………i
`implies2y(dy)/(dx)=5implies(dy)/(dx)=5/(2y)`
At `(1,-2), (dy)/(dx)=(5/(2y))_((1,-2))=(-5)/4`
Slope of normal `=(-1)/(-5//4)=4/5`
`:.` Equation of normal at the point `(1,-2)` is
`y+2=4/5(x-1)`
`implies5y+10=4x-4`
`implies4x-5y-14=0`.................ii
As the normal is of the form
`ax-5y+b=0`
On comparing this with Eq. ii we get
`a=4` and `b=-14`