Posted by **rob** on Friday, July 22, 2011 at 8:27am.

the tangent yo the curve y=x^2 +5x -2 @ the point (1,4)intersect the normal to the same curve @ the point (-3,-8) at the point P.Find the coordinates of point P.[ans: -1/3,-16/3]

just give me some hint to calculate this solution.

- math -
**Reiny**, Friday, July 22, 2011 at 8:50am
dy/dx = 2x + 5

slope of tangent at (1,4) = 2(1) +5 = 7

at (1,4) ...

4 = 7(1) + b

b= -3

tangent equation at (1,4) is y = 7x - 3

at (-3,-8)

slope of tangent is 2(-3) + 5 = -1

so slope of normal is +1

equation of normal is y = x + b

at (-3,-8) , -8 = -3 + b

b = -5

equation of normal is y = x - 5

where do they intersect ?

when 7x - 3 = x - 5

6x = -2

x = -1/3

back into y = x-5 , (or the other equation, makes no differencen)

y = - 1/3 - 5 = -1/3 - 15/3 = -16/3

so the two straight lines intersect at (-1/3, -16/3)

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