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)
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.
1 answer