Problem Statement
Consider the curve defined by $x^3 + y^3 = 6xy$.
(a) Show that $\frac{dy}{dx} = \frac{2y - x^2}{y^2 - 2x}$. [3 marks]
(b) Find the equation of the normal to the curve at the point $(3, 3)$. [3 marks]
Verified Solution & Marking Scheme
Part (a): Differentiate Implicitly
$\frac{d}{dx}(x^3 + y^3) = \frac{d}{dx}(6xy)$
$3x^2 + 3y^2 \frac{dy}{dx} = 6 \left( y + x \frac{dy}{dx} \right) = 6y + 6x \frac{dy}{dx}$
Divide through by 3:
$x^2 + y^2 \frac{dy}{dx} = 2y + 2x \frac{dy}{dx}$
$\frac{dy}{dx}(y^2 - 2x) = 2y - x^2 \implies \frac{dy}{dx} = \frac{2y - x^2}{y^2 - 2x}$
Part (b): Gradient and Normal at (3, 3)
At $(3, 3)$:
$m_{\text{tangent}} = \frac{2(3) - 3^2}{3^2 - 2(3)} = \frac{6 - 9}{9 - 6} = \frac{-3}{3} = -1$
Gradient of normal $m_{\text{normal}} = -\frac{1}{-1} = 1$.
Equation of Normal
$y - 3 = 1(x - 3) \implies y = x$