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Difficulty: HardTangents and Normals to Curves

The line y=5x+cy = 5x + c is tangent to the curve y=x3−7x+15y = x^3 - 7x + 15 at a point PP located in the first quadrant. What is the equation of the normal line to the curve at point PP?

  1. x+5y−47=0x + 5y - 47 = 0Answer
  2. B
    5x−y−1=05x - y - 1 = 0
  3. C
    x+5y−43=0x + 5y - 43 = 0
  4. D
    5x+y−19=05x + y - 19 = 0

Answer

The equation of the normal line to the curve at point P is x+5y−47=0x + 5y - 47 = 0.
Differentiating the curve yields dydx=3x2−7\frac{dy}{dx} = 3x^2 - 7. Equating this derivative to the tangent slope of 5 gives x=2x = 2 for the first quadrant. Evaluating the curve equation at x=2x = 2 gives y=9y = 9, locating point P(2,9)P(2, 9). The normal gradient is the negative reciprocal of the tangent slope, giving mn=−15m_n = -\frac{1}{5}. Applying the point-slope formula with (2,9)(2, 9) yields y−9=−15(x−2)y - 9 = -\frac{1}{5}(x - 2), which simplifies to x+5y−47=0x + 5y - 47 = 0.

Step-by-Step Solution

1
Find the x-coordinate of the point of tangency P
x=2x = 2
The gradient of the curve dydx=3x2−7\frac{dy}{dx} = 3x^2 - 7 must equal the gradient of the given line m=5m = 5. Solving 3x2−7=53x^2 - 7 = 5 gives 3x2=12  ⟹  x2=4  ⟹  x=23x^2 = 12 \implies x^2 = 4 \implies x = 2 (since PP is in the first quadrant).
2
Calculate the y-coordinate of point P
y=9y = 9
Substitute x=2x = 2 into the curve equation y=23−7(2)+15=8−14+15=9y = 2^3 - 7(2) + 15 = 8 - 14 + 15 = 9. Thus, P=(2,9)P = (2, 9).
3
Determine the gradient of the normal line
mn=−15m_n = -\frac{1}{5}
The normal line is perpendicular to the tangent line. Since the tangent gradient mt=5m_t = 5, the normal gradient is mn=−1mt=−15m_n = -\frac{1}{m_t} = -\frac{1}{5}.
4
Formulate the equation of the normal line
x+5y−47=0x + 5y - 47 = 0
Using the point-slope equation y−y1=mn(x−x1)y - y_1 = m_n(x - x_1) with point (2,9)(2, 9) and gradient −15-\frac{1}{5}: y−9=−15(x−2)  ⟹  5(y−9)=−(x−2)  ⟹  5y−45=−x+2  ⟹  x+5y−47=0y - 9 = -\frac{1}{5}(x - 2) \implies 5(y - 9) = -(x - 2) \implies 5y - 45 = -x + 2 \implies x + 5y - 47 = 0.

Key Concept

The gradient of the normal to a curve at a given point is the negative reciprocal of the derivative (tangent gradient) at that point: mnormal=−1dydxm_{normal} = -\frac{1}{\frac{dy}{dx}}.
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Topics

Tangents and Normals to Curves

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