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Nov 2016 p13 q10
1377
A curve is such that \(\frac{dy}{dx} = \frac{2}{a}x^{-\frac{1}{2}} + ax^{-\frac{3}{2}}\), where \(a\) is a positive constant. The point \(A(a^2, 3)\) lies on the curve. Find, in terms of \(a\),
the equation of the tangent to the curve at \(A\), simplifying your answer,
the equation of the curve.
It is now given that \(B(16, 8)\) also lies on the curve.
Find the value of \(a\) and, using this value, find the distance \(AB\).
Solution
(i) At \(x = a^2\), \(\frac{dy}{dx} = \frac{2}{a}a^{-1} + a \cdot a^{-3} = \frac{3}{a^2}\). The equation of the tangent is \(y - 3 = \frac{3}{a^2}(x - a^2)\), simplifying to \(y = \frac{3}{a^2}x\).
(ii) Integrate \(\frac{dy}{dx} = \frac{2}{a}x^{-\frac{1}{2}} + ax^{-\frac{3}{2}}\) to get \(y = \frac{2}{a} \cdot \frac{x^{\frac{1}{2}}}{\frac{1}{2}} + a \cdot \frac{x^{-\frac{1}{2}}}{-\frac{1}{2}} + c = \frac{4x^{\frac{1}{2}}}{a} - 2ax^{-\frac{1}{2}} + c\). Substituting \(x = a^2, y = 3\) gives \(c = 1\), so \(y = \frac{4x^{\frac{1}{2}}}{a} - 2ax^{-\frac{1}{2}} + 1\).
