DIfferentiation

Question

The tangent and normal to the curve y=4√(x+2) at the point P(7,12) cut the x axis at M and N respectively.Calculate the area of the triangle PMN.

Answer 1

Rewrite the function as y = 4(x+2)^1/2

so y' = 2(x + 2)^-1/2  So y'(7) = 2*(7+2)^-1/2 = 2/3 = slope of tangent at point P.

 

Equation of tangent is y-12 = 2/3(x-7). The y-intercept, N, = f(0), so y-12 = 2/3(0-7), y=22/3

N is the point (0, 22/3)

 

The slope of the normal must be -1/(2/3) = -3/2

Equation of the normal is: y-12 = -3/2(x-7)

The x-intercept, M, occurs where y = 0, so 0-12 = -3/2(x-7), so x = 15. Point M = (15, 0)

 

We now calculate PN and PM using the distance formula d = √[(x₁ - x₂)² + (y₁ - y₂)²]

 

PN = √(637) /3 and PM = 2√52

 

Area of right triangle = 1/2√(637) /3 * 2√52 = √(33124) /3 = 182/3 ANSWER