(a) Find the equation of the tangent to curve (frac{x^{2}}{4} + y^{2} = 1) at point (1, frac{sqrt{3}}{2}).
(b) Express (frac{3x + 2}{x^{2} + x – 2}) in partial fractions.
Explanation
(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}{2sqrt{3}})
Equation of the tangent at ((1, frac{sqrt{3}}{2})) is
(frac{y – frac{sqrt{3}}{2}}{x – 1} = -frac{1}{2sqrt{3}})
(1 – x = 2sqrt{3} (y – frac{sqrt{3}}{2}))
(1 – x = 2sqrt{3} y – 3 implies 2sqrt{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)}).