The tangent to `y=a x^2+b x+7/2a t(1,2)` is parallel to the normal at the point `(-2,2)` on the curve `y=x^2+6x+10 ,` then `a=-1` b. `a=1` c. `b=5/2` d. `b=-5/2`

Options

A. `a = 1`
B. `a = -1`
C. `b = (-15)/(2)`
D. `b = (5)/(2)`

Answer

A

Explanation

`y = x^(2) + 6x + 10`
`:. (dy)/(dx) = 2x + 6`
`(dy)/(dx)|(-2,2) = 2`
`:.` slope of Normal,`m = - 1//2`
`:. y = ax^(2) + bx + (7)/(2)`
Passes through `(1,2) rArr a+ b = -3//2`…..(1)
Also `(dy)/(dx)|(1,2) = 2a + b`
`rArr - (1)/(2) = 2a + b`......(2)
`[1]` and [2] `rArr a = 1 , b = -5//2`