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The tangent to y=a x^2+b x+7/2a t(1,2) i...

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`

A

a=1

B

a=-1

C

b=-5/2

D

b=5/2

Text Solution

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The correct Answer is:
A, C
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