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2019 Paper 1 Q1
D: 1500.0 B: 1500.0
differentiation optimisation trigonometric triangle area perimeter stationary points coordinate geometry

A straight line passes through the fixed point \((1 , k)\) and has gradient \(- \tan \theta\), where \(k > 0\) and \(0 < \theta < \frac{1}{2}\pi\). Find, in terms of \(\theta\) and \(k\), the coordinates of the points \(X\) and \(Y\) where the line meets the \(x\)-axis and the \(y\)-axis respectively.

  1. Find an expression for the area \(A\) of triangle \(OXY\) in terms of \(k\) and \(\theta\). (The point \(O\) is the origin.) You are given that, as \(\theta\) varies, \(A\) has a minimum value. Find an expression in terms of \(k\) for this minimum value.
  2. Show that the length \(L\) of the perimeter of triangle \(OXY\) is given by $$L = 1 + \tan \theta + \sec \theta + k(1 + \cot \theta + \cosec \theta).$$ You are given that, as \(\theta\) varies, \(L\) has a minimum value. Show that this minimum value occurs when \(\theta = \alpha\) where $$\frac{1 - \cos \alpha}{1 - \sin \alpha} = k.$$ Find and simplify an expression for the minimum value of \(L\) in terms of \(\alpha\).

Solution: \(y = (-\tan \theta)(x-1)+k\) so when \(x = 0\), \(y = k + \tan \theta\), so \(Y = (0, k+\tan \theta)\). When \(y = 0\), \(x = 1 + \frac{k}{\tan \theta}\)

  1. \(A = \frac12 (k+\tan \theta)\left ( 1 + \frac{k}{\tan \theta} \right) = k + \frac12 \left (\tan \theta + \frac{k^2}{\tan \theta} \right)\) Notice that \(x + \frac{k^2}{x} \geq 2 k\) by AM-GM, so the minimum is \(k + \frac12 \cdot 2k = 2k\)
  2. \(\,\) \begin{align*} L &= k + \tan \theta + 1 + k \cot \theta + \sqrt{(k + \tan \theta)^2 + \left (1 + \frac{k}{\tan \theta} \right)^2} \\ &= k + \tan \theta + 1 + k \cot \theta + \sqrt{k^2 + 2 k \tan \theta +\tan^2 \theta + 1 + 2k \cot \theta + k^2\cot^2 \theta} \\ &= k + \tan \theta + 1 + k \cot \theta + \sqrt{\sec^2 \theta+ 2k \sec\theta\cosec \theta + k^2\cosec^2 \theta} \\ &= k + \tan \theta + 1 + k \cot \theta +\sec \theta + k\cosec \theta\\ &= 1 + \tan \theta + \sec \theta + k (1 + \cot \theta + \cosec \theta) \end{align*} \begin{align*} && \frac{\d L}{\d \theta} &= \sec^2 \theta + \tan \theta \sec \theta + k(-\cosec^2 \theta - \cot \theta \cosec \theta ) \\ \Rightarrow && 0 &=\sec^2 \alpha+ \tan \theta \sec \alpha+ k(-\cosec^2 \alpha- \cot \alpha\cosec \alpha) \\ \Rightarrow && k &= \frac{\sec^2 \alpha+ \tan \alpha\sec \alpha}{\cosec^2 \alpha+ \cot \alpha\cosec \alpha} \\ &&&= \frac{\sin^2 \alpha(1 + \sin \alpha)}{\cos^2 \alpha (1+ \cos \alpha)} \\ &&&= \frac{(1-\cos^2 \alpha)(1 + \sin \alpha)}{(1-\sin^2 \alpha )(1+ \cos \alpha)} \\ &&&= \frac{1-\cos \alpha}{1-\sin \alpha} \\ \Rightarrow && L &= 1 + \tan \alpha + \sec \alpha + \frac{1-\cos \alpha}{1-\sin \alpha} \left (1 + \cot \alpha + \cosec \alpha \right) \\ &&&= \frac{1+\tan \alpha + \sec \alpha -\sin \alpha-\sin \alpha \tan \alpha-\tan \alpha}{1-\sin \alpha} + \\ &&&\quad \quad \frac{1+\cot \alpha + \cosec \alpha-\cos \alpha-\cos \alpha \cot \alpha -\cot \alpha}{1-\sin \alpha} \\ &&&= \frac{2+\sec \alpha(1-\sin^2 \alpha)-\sin \alpha + \cosec \alpha(1-\cos^2 \alpha)-\cos \alpha}{1-\sin \alpha} \\ &&&= \frac{2+\cos\alpha-\sin \alpha + \sin\alpha-\cos \alpha}{1-\sin \alpha} \\ &&&= \frac{2}{1-\sin \alpha} \end{align*}

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