System of Linear Equations: Curve coefficients from geometric conditions — RI 2025 H2 Math Prelim Paper 1
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Question
A curve has equation \(y = ax + b + \dfrac{c}{x^2 - 1}\), where \(a\), \(b\) and \(c\) are real constants. It is given that the curve crosses the \(x\)-axis at \(x = 2\). The normal to the curve at the point \((0, 3)\) meets the \(x\)-axis at \(x = 7.5\). Find the values of \(a\), \(b\) and \(c\).
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A curve has equation \(y = ax + b + \dfrac{c}{x^2-1}\), where \(a\), \(b\) and \(c\) are real constants. It crosses the \(x\)-axis at \(x = 2\). The normal to the curve at \((0,3)\) meets the \(x\)-axis at \(x = 7.5\). Find the values of \(a\), \(b\) and \(c\).
Since \((2,0)\) and \((0,3)\) lie on the curve, \[\begin{aligned} 2a + b + \tfrac{1}{3}c &= 0 \quad\cdots(1)\\ b - c &= 3 \quad\cdots(2) \end{aligned}\]
Differentiating, \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = a - \dfrac{2cx}{(x^2-1)^2}\).
Gradient of normal at \((0,3)\) is \(\dfrac{3-0}{0-7.5} = -\dfrac{2}{5}\).
So gradient of tangent at \((0,3)\) is \(\dfrac{5}{2}\), giving \[ a = \frac{5}{2} \quad\cdots(3) \] Solving (1), (2) and (3) simultaneously (using GC), \[ a = \frac{5}{2},\quad b = -3,\quad c = -6. \]