The slope of normal to be parabola `y = (x^(2))/(4) -2` drawn through the point `(10,-1)` is (a) -2 (b) `-sqrt(3)` (c) `-1/2` (d) `-5/3`

Options

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

Answer

C

Explanation

`x^(2) = 4(y+2)` is the given parabola.
Any normal is `x = m (y+2) -2m -m^(3)`
If `(10,-1)` lies on this line then
`10 = m -2m -m^(3)`
`rArrm^(3) + m + 10 =0 rArr m =-2`
Slope of normal `= 1//m =- 1//2`.