WAEC Further Mathematics 2018 Paper

We are given the equation $x^{2} + y^{2} - 4x - 2y = 0$

$y = x^{2} + y^{2} - 4x - 2y$

Using the method of implicit differentiation,

$\frac{dy}{dx} = 2x + 2y\frac{dy}{dx} - 4 - 2\frac{dy}{dx}$

For the tangent, $\frac{dy}{dx} = 0$,

$\therefore 2x + 2y\frac{dy}{dx} - 4 - 2\frac{dy}{dx} = 0$

$(2y - 2)\frac{dy}{dx} = 4 - 2x \implies \frac{dy}{dx} = \frac{4 - 2x}{2y - 2}$

At (1, 3), $\frac{dy}{dx} = \frac{4 - 2(1)}{2(3) - 2} = \frac{2}{4} = \frac{1}{2}$

Equation: $\frac{y - 3}{x - 1} = \frac{1}{2} \implies 2y - 6 = x - 1$

= $2y - x - 6 + 1 = 2y - x - 5 = 0$