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The slope of normal to be parabola y = (...

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`

Text Solution

Verified by Experts

The correct Answer is:
C
`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`.
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