A curve $C$ has equation $x^3+y^3-5xy=0$ where $x0$. Find \(\frac{\mathrm{d}y}{\mathr

Question

A curve $C$ has equation $x^3+y^3-5xy=0$ where $x>0$. Find $\frac{\mathrm{d}y}{\mathrm{d}x}$ in terms of $x$ and $y$.[2] Find the coordinates of the point on $C$ at which the normal is parallel to the $y$-axis.[3] Determine the nature of the stationary point.[2]

Accepted Answer

(a) $\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{5y-3x^2}{3y^2-5x}$ (b) $\left(\dfrac{5\sqrt[3]2}{3},\dfrac{5\sqrt[3]4}{3}\right)$ (c) Maximum point

2025 DHS P2 Q3

2025 DHS P2 Q3

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