WAEC 2011 · Paper 2 · Q12

(a)
Find the equation of the tangent to the curve $\dfrac{x^2}{4} + y^2 = 1$ at the point $\left(1, \dfrac{\sqrt3}{2}\right)$ .
(b)
Express $\dfrac{3x + 2}{x^2 + x - 2}$ in partial fractions.

Worked solution (try it first)

Differentiate each term with respect to $x$ .
$y^2$ gives $2y\dfrac{dy}{dx}$ : $\dfrac{x}{2} + 2y\dfrac{dy}{dx} = 0$ .
Make $\dfrac{dy}{dx}$ the subject: $\dfrac{dy}{dx} = -\dfrac{x}{4y}$ .
At $\left(1, \frac{\sqrt3}{2}\right)$ : $\dfrac{dy}{dx} = -\dfrac{1}{4 \times \frac{\sqrt3}{2}}$
$= -\dfrac{1}{2\sqrt3}$ .
The tangent: $y - \dfrac{\sqrt3}{2} = -\dfrac{1}{2\sqrt3}(x - 1)$ .
Multiply by $2\sqrt3$ : $2\sqrt3y - 3 = -x + 1$ .
Rearrange: $x + 2\sqrt3y - 4 = 0$ .

Factorise the bottom: $x^2 + x - 2 = (x + 2)(x - 1)$ .
Write $\dfrac{A}{x + 2} + \dfrac{B}{x - 1}$ and multiply through: $3x + 2 = A(x - 1) + B(x + 2)$ .
Put $x = 1$ : $5 = 3B$ , so $B = \frac53$ .
Put $x = -2$ : $-4 = -3A$ , so $A = \frac43$ .
So the answer is $\dfrac{4}{3(x + 2)} + \dfrac{5}{3(x - 1)}$ .