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9709 P11 - Jun 2013 - Q10
1297
The diagram shows part of the curve \(y = (x - 2)^4\) and the point \(A (1, 1)\) on the curve. The tangent at \(A\) cuts the \(x\)-axis at \(B\) and the normal at \(A\) cuts the \(y\)-axis at \(C\).
Find the coordinates of \(B\) and \(C\).
Find the distance \(AC\), giving your answer in the form \(\frac{\sqrt{a}}{b}\), where \(a\) and \(b\) are integers.
Find the area of the shaded region.
Solution
(i) Differentiate \(y = (x - 2)^4\) to find the gradient of the tangent: \(\frac{dy}{dx} = 4(x - 2)^3\). At \(x = 1\), the gradient is \(-4\). The equation of the tangent is \(y - 1 = -4(x - 1)\), which gives \(B \left( \frac{5}{4}, 0 \right)\). The gradient of the normal is \(\frac{1}{4}\), so the equation of the normal is \(y - 1 = \frac{1}{4}(x - 1)\), which gives \(C \left( 0, \frac{3}{4} \right)\).
(ii) Use the distance formula: \(AC^2 = 1^2 + \left( \frac{1}{4} \right)^2\), so \(AC = \frac{\sqrt{17}}{4}\).
(iii) Integrate \((x - 2)^4\) from 1 to 2: \(\int (x - 2)^4 \, dx = \frac{(x - 2)^5}{5}\). Evaluate from 1 to 2 to get \(\frac{1}{5}\). The area of the triangle formed by the tangent and normal is \(\frac{1}{2} \times 1 \times \frac{5}{4} = \frac{5}{8}\). The area of the shaded region is \(\frac{5}{8} - \frac{1}{5} = \frac{3}{40}\) or \(0.075\).
