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Zorluk: ZorTangents and Normals to Curves

The line 3x−y+k=03x - y + k = 0 is a normal to the curve y=x+2x−1y = \frac{x + 2}{x - 1} at a point PP located in the first quadrant. What is the value of the constant kk?

  1. -10Cevap
  2. B
    -2
  3. C
    10
  4. D
    6

Cevap

The value of the constant kk is −10-10.
Differentiating y=x+2x−1y = \frac{x + 2}{x - 1} yields dydx=−3(x−1)2\frac{dy}{dx} = \frac{-3}{(x - 1)^2}. The gradient of the normal line is mN=−1dy/dx=(x−1)23m_N = -\frac{1}{dy/dx} = \frac{(x - 1)^2}{3}. Equating mNm_N to the gradient of 3x−y+k=03x - y + k = 0 (which is 33) gives (x−1)23=3  ⟹  (x−1)2=9\frac{(x - 1)^2}{3} = 3 \implies (x - 1)^2 = 9. Solving gives x=4x = 4 or x=−2x = -2. Since point PP lies in the first quadrant, x=4x = 4, which gives y=4+24−1=2y = \frac{4 + 2}{4 - 1} = 2. Substituting (4,2)(4, 2) into 3x−y+k=03x - y + k = 0 gives 3(4)−2+k=03(4) - 2 + k = 0, so k=−10k = -10.

Adım Adım Çözüm

1
Differentiate the curve equation to find the tangent gradient function.
Using the quotient rule on y=x+2x−1y = \frac{x + 2}{x - 1}, dydx=1(x−1)−1(x+2)(x−1)2=−3(x−1)2\frac{dy}{dx} = \frac{1(x - 1) - 1(x + 2)}{(x - 1)^2} = \frac{-3}{(x - 1)^2}.
The derivative gives the gradient of the tangent line to the curve at any point xx.
2
Determine the gradient function for the normal line.
The normal gradient is mN=−1dy/dx=(x−1)23m_N = -\frac{1}{dy/dx} = \frac{(x - 1)^2}{3}.
The normal line is perpendicular to the tangent line, so its gradient is the negative reciprocal of dydx\frac{dy}{dx}.
3
Equate the normal gradient to the gradient of the given line and solve for xx.
The line 3x−y+k=03x - y + k = 0 has gradient 33. Setting (x−1)23=3  ⟹  (x−1)2=9  ⟹  x−1=±3\frac{(x - 1)^2}{3} = 3 \implies (x - 1)^2 = 9 \implies x - 1 = \pm 3, giving x=4x = 4 or x=−2x = -2.
The normal line at point PP must be parallel to (and thus have the same gradient as) the given line 3x−y+k=03x - y + k = 0.
4
Select the coordinate corresponding to the first quadrant and calculate the yy-coordinate.
Since PP lies in the first quadrant (x>0,y>0x > 0, y > 0), x=4x = 4. Substituting x=4x = 4 into the curve equation gives y=4+24−1=2y = \frac{4 + 2}{4 - 1} = 2. Thus, P=(4,2)P = (4, 2).
The problem specifies that point PP is located in the first quadrant.
5
Substitute point P(4,2)P(4, 2) into the line equation to solve for kk.
Substituting x=4x = 4 and y=2y = 2 into 3x−y+k=03x - y + k = 0 yields 3(4)−2+k=0  ⟹  10+k=0  ⟹  k=−103(4) - 2 + k = 0 \implies 10 + k = 0 \implies k = -10.
Point PP lies on the normal line, so its coordinates must satisfy the line equation.

Anahtar Kavram

The normal line to a curve at a point PP is perpendicular to the tangent line at PP, with gradient mnormal=−1dy/dxm_{\text{normal}} = -\frac{1}{dy/dx}. A point PP on the curve must satisfy both the curve equation and the normal line equation.
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Tangents and Normals to Curves

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