關於Differentiation的A.Maths.問題2

y = 2x2 + x – 1
dy/dx = 4x + 1
Slope of tangent at the point (1,2) = 5
Slope of normal at the point (1,2) = -1/5
Equation of the normal is (y – 2) / (x – 1) = - 1/5
5y – 10 = 1 – x
5y + x – 11 = 0
The y-intercept of the normal = 11/5 = 2.2
The x-intercept of the normal = 11
Area of the triangle = 11 * 2.2 / 2 = 12.1

diff the curve w.r.t.x
dy/dx = 4x+1
dy/dx (1,2) = 4(1)+1 = 5
The slope of the normal line = -1/5
The equation of the normal line : (y- 2)/(x-1) = -1/5
5(y-2) = -x+1
5y + x - 11 = 0
=> y = (11- x)/5
x-intercept : y= 0, x = 11
y-intercept : x = 0, y = 11/5
area of triangle =∫(11-x)/5 dx (from x = 0 to x = 11)
= [11x - x^2/2]/5 (from x = 0 to x = 11)
= [11(11) - 11^2/2]/5 - 0
= 121/10
= 12.1 sq. unit