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The curve C has equation $$y = 9 - 4x - \frac{8}{x}, \; x > 0.$$ The point P on C has x-coordinate equal to 2 - Edexcel - A-Level Maths Pure - Question 2 - 2008 - Paper 1

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The-curve-C-has-equation--$$y-=-9---4x---\frac{8}{x},-\;-x->-0.$$---The-point-P-on-C-has-x-coordinate-equal-to-2-Edexcel-A-Level Maths Pure-Question 2-2008-Paper 1.png

The curve C has equation $$y = 9 - 4x - \frac{8}{x}, \; x > 0.$$ The point P on C has x-coordinate equal to 2. (a) Show that the equation of the tangent to C a... show full transcript

Worked Solution & Example Answer:The curve C has equation $$y = 9 - 4x - \frac{8}{x}, \; x > 0.$$ The point P on C has x-coordinate equal to 2 - Edexcel - A-Level Maths Pure - Question 2 - 2008 - Paper 1

Step 1

Show that the equation of the tangent to C at the point P is $y = 1 - 2x$

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Answer

  1. Find the derivative of the function:
    To find the slope of the tangent, first compute the derivative of the curve:
    dydx=−4+8x2\frac{dy}{dx} = -4 + \frac{8}{x^{2}}

  2. Evaluate the derivative at the point P:
    At x=2x = 2, we have:
    dydx=−4+822=−4+2=−2.\frac{dy}{dx} = -4 + \frac{8}{2^{2}} = -4 + 2 = -2.

  3. Find the y-coordinate of point P:
    Substitute x=2x = 2 into the equation of the curve:
    y=9−4(2)−82=9−8−4=−3.y = 9 - 4(2) - \frac{8}{2} = 9 - 8 - 4 = -3.
    Thus, point P is (2,−3)(2, -3).

  4. Write the equation of the tangent line:
    Using the point-slope form of the line:
    y−y1=m(x−x1)y - y_{1} = m(x - x_{1})
    where m=−2m = -2 and (x1,y1)=(2,−3)(x_{1}, y_{1}) = (2, -3), we get:
    y+3=−2(x−2).y + 3 = -2(x - 2).
    Simplifying this yields:
    y=−2x+4−3=−2x+1,y = -2x + 4 - 3 = -2x + 1,
    or equivalently:
    y=1−2x.y = 1 - 2x.

Step 2

Find an equation of the normal to C at the point P.

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Answer

  1. Determine the gradient of the normal line:
    The gradient of the normal is the negative reciprocal of the tangent's gradient.
    Given the tangent's gradient at P is −2-2, the gradient of the normal is:
    mnormal=−1−2=12.m_{normal} = -\frac{1}{-2} = \frac{1}{2}.

  2. Use point-slope form to write the normal's equation:
    Again, we apply the point-slope formula using point P (2,−3)(2, -3):
    y−(−3)=12(x−2).y - (-3) = \frac{1}{2}(x - 2).

  3. Simplify the equation:
    Rearranging gives us:
    y+3=12x−1,y + 3 = \frac{1}{2}x - 1,
    or equivalently:
    y=12x−4.y = \frac{1}{2}x - 4.

Step 3

Find the area of triangle APB.

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Answer

  1. Find the coordinates of points A and B:
    • The tangent intersects the x-axis at A when y=0y = 0:
      0=1−2x⇒2x=1⇒x=0.5⇒A(0.5,0).0 = 1 - 2x \Rightarrow 2x = 1 \Rightarrow x = 0.5 \Rightarrow A(0.5, 0).
    • The normal intersects the x-axis at B when y=0y = 0:
      0=12x−4⇒12x=4⇒x=8⇒B(8,0).0 = \frac{1}{2}x - 4 \Rightarrow \frac{1}{2}x = 4 \Rightarrow x = 8 \Rightarrow B(8, 0).
  2. Calculate the area of triangle APB:
    The base length AB is ∣0.5−8∣=7.5|0.5 - 8| = 7.5, and the height (from P to the x-axis) is ∣yP∣=3|y_{P}| = 3.
    Thus, the area of triangle APB is:
    Area=12×base×height=12×7.5×3=11.25.\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 7.5 \times 3 = 11.25.

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