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2011 Paper 1 Q1
D: 1500.0 B: 1479.0
implicit differentiation gradient curve and line intersection tangent-normal simultaneous equations algebraic manipulation

  1. Show that the gradient of the curve \(\; \dfrac a x + \dfrac by =1\), where \(b\ne0\), is \(\; -\dfrac{ay^2}{bx^2}\,\). The point \((p,q)\) lies on both the straight line \(ax+by=1\) and the curve \(\dfrac a x + \dfrac by =1\,\), where \(ab\ne0\). Given that, at this point, the line and the curve have the same gradient, show that \( p=\pm q\,\). Show further that either \((a-b)^2 =1\,\) or \((a+b)^2 =1\,\).
  2. Show that if the straight line \(ax+by=1\), where \(ab\ne0\), is a normal to the curve \(\dfrac a x - \dfrac by =1\), then \(a^2-b^2 = \frac12\,\).

Solution:

  1. \(\,\) \begin{align*} && 1 &= \frac{a}{x} + \frac{b}{y} \\ \frac{\d}{\d x}: && 0 &= -\frac{a}{x^2} - \frac{b}{y^2} \frac{\d y}{\d x} \\ \Rightarrow && \frac{\d y}{\d x} &= -\frac{ay^2}{bx^2} \\ \\ (p,q): && -\frac{aq^2}{bp^2} &= -\frac{a}{b} \\ \Rightarrow && p^2 &= q^2 \\ \Rightarrow && p &= \pm q \\ \\ \Rightarrow && ap \pm b p &= 1 \\ \Rightarrow && (a\pm b)p &= 1 \\ \Rightarrow && \frac{a}{p} \pm \frac{b}{p} &= 1 \\ \Rightarrow && (a \pm b)\frac{1}{p} &= 1 \\ \Rightarrow && (a \pm b)^2 &= 1 \end{align*}
  2. \(\,\) \begin{align*} && 1 &= \frac{a}{x} - \frac{b}{y} \\ \Rightarrow && \frac{\d y}{\d x} &= \frac{ay^2}{bx^2} \\ \Rightarrow && \frac{aq^2}{bp^2} &= \frac{b}{a} \\ \Rightarrow && aq &= \pm bp \\ \Rightarrow && 1 &= \frac{a}{p} - \frac{b}{q} \\ &&&= \frac{aq-bp}{pq} \\ \Rightarrow && aq &= -bp \\ \Rightarrow && 1 &= \frac{2aq}{pq} \\ \Rightarrow && p &= 2a \\ \Rightarrow && q &= -2b \\ \Rightarrow && 1 &= 2a^2-2b^2 \\ \Rightarrow && \frac12 &= a^2-b^2 \end{align*}

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