曲線與直線相截微分題

A straight line y=k cuts the curve C: y = x² + 2x at two points A and B,where k is a constant. The tangents to the curve C at A and B are perpendicularto each other.
(a) Find the value of k
(b) Find the equations of the tangents to curve C at A and B.

Solution :
(a)
The straight line : y = k ...... [1]
The curve C : y = x² + 2x ...... [2]

[1] = [2] :
x² + 2x = k
x² + 2x - k = 0
x = -1 + √(1 + k) or x = -1 - √(1 + k)
A = (-1 + √(1 + k), k), B = (-1 -√(1 + k), k)

C : y = x² + 2x
y' = 2x + 2

Slope of the tangent at A = 2[-1 + √(1 + k)] + 2 = 2√(1 + k)
Slope of the tangent at A = 2[-1 - √(1 + k)] + 2 = -2√(1 + k)

The two tangents are perpendicular to each other :
[2√(1 + k)] ´ [-2√(1 + k)] = -1
4(1 + k) = 1
4 + 4k = 1
4k = -3
k = -3/4

(b)
Slope of the tangent at A = 2√[1 + (-3/4)] = 1
A = (-1 + √[1 + (-3/4)], -3/4) = (-1/2,-3/4)
Equation of the tangent at A :
y + (3/4) = 1[x + (1/2)]
x - y - (1/4) = 0
4x - 4y - 1 = 0

Slope of the tangent at B = -2√[1 + (-3/4)] = -1
B = (-1 - √[1 + (-3/4)], -3/4) = (-3/2,-3/4)
Equation of the tangent at B :
y + (3/4) = (-1)[x + (3/2)]
x + y + (9/4) = 0
4x + 4y + 9 = 0

The two required equations are :
4x - 4y - 1 = 0 and 4x + 4y + 9 = 0

y = x^2 + 2x
dy/dx = 2x + 2
y = k .................(1)
y = x^2 + 2x .........(2)
so x^2 + 2x - k = 0
Let the 2 roots of the equation be a and b
so slope at x = a is (2a + 2)
slope at x = b is (2b + 2)
since the 2 lines are perpendicular to each other,
(2a + 2)(2b + 2) = - 1
4ab + 4(a + b) + 5 = 0
sum of roots = a + b = - 2, product of roots = ab = - k
that is - 4k + 4(-2) + 5 = 0
- 4k - 3 = 0
so k = -3/4.
Sub k into the quadratic equation
x^2 + 2x + 3/4 = 0
4x^2 + 8x + 3 = 0
(2x + 1)(2x + 3) = 0
x = -1/2 and -3/2
so the 2 points are A(-1/2, -3/4) and (-3/2, -3/4).
Slope at A = 2(-1/2) + 2 = 1
Slope at B = 2(-3/2) + 2 = - 1
So the 2 tangents are :
y + 3/4 = x + 1/2 and
y + 3/4 = - x - 3/2
that is 4y + 3 = 4x + 2 or 4x - 4y - 1 = 0 and
4y + 3 = - 4x - 6 or 4x + 4y + 9 = 0