(a) \(\frac{x^{2}}{4} + y^{2} = 1\)
Differentiating w.r.t x,
\(\frac{\mathrm d y}{\mathrm d x} = -\frac{2x}{4(2y)} = -\frac{x}{4y}\)
The gradient of the curve is
\(\frac{\mathrm d y}{\mathrm d x} = -\frac{1}{4(\frac{\sqrt{3}}{2})}\)
= \(-\frac{1}{2\sqrt{3}}\)
Equation of the tangent at \((1, \frac{\sqrt{3}}{2})\) is
\(\frac{y - \frac{\sqrt{3}}{2}}{x - 1} = -\frac{1}{2\sqrt{3}}\)
\(1 - x = 2\sqrt{3} (y - \frac{\sqrt{3}}{2})\)
\(1 - x = 2\sqrt{3} y - 3 \implies 2\sqrt{3} y + x - 4 = 0\).
(b) \(\frac{3x + 2}{x^{2} + x - 2}\)
\(x^{2} + x - 2 = x^{2} - x + 2x - 2\)
\(x(x - 1) + 2(x - 1) \implies x^{2} + x - 2 \equiv (x - 1)(x + 2)\)
\(\frac{3x + 2}{x^{2} + x - 2} = \frac{A}{x - 1} + \frac{B}{x + 2}\)
= \(\frac{A(x + 2) + B(x - 1)}{(x - 1)(x + 2)}\)
Comparing with the equation given, we have
\(3x + 2 = A(x + 2) + B(x - 1)\)
= \(Ax + 2A + Bx - B\)
\(\implies A + B = 3 ... (1)\)
\(2A - B = 2 ... (2)\)
(1) + (2) : \(3A = 5 \implies A = \frac{5}{3}\)
\(\frac{5}{3} + B = 3 \implies B = 3 - \frac{5}{3} = \frac{4}{3}\)
\(\therefore \frac{3x + 2}{x^{2} + x - 2} = \frac{5}{3(x - 1)} + \frac{4}{3(x + 2)}\).