math
posted by rob .
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.

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 = x5 , (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)