the tangent to the curve y= ax^2 + bx 2 at (1,1/2) is parallel to the normal to the curve y= x^2 +6x +10 at (-2,2). find the values of a and b
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Find the slope of y= x^2 +6x +10 at (-2,2):
y' = 2x + 1 so slope is -3
The normal line will have slope 1/3 (negative reciprocal).
So the slope to ax^2 + bx +? 2 at (1, 1/2) is 1/3.
Plug in the point (1, 1/2) into the function to get one condition on a and b.
Plug in the point x=1 and m=1/3 into the derivative of the curve.
Then solve for a and b.
Good luck!!