The curve $C$ has equation ${{y}^{3}}=4-\frac{x{{y}^{2}}}{2}$. Show that $\frac{\mathrm{d}y}{\mathrm

Question

The curve $C$ has equation ${{y}^{3}}=4-\frac{x{{y}^{2}}}{2}$. Show that $\frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{y}{6y+2x}$. [2] Find the equation of the normal to $C$ at the point $P$ where $y=1$. [3]

Accepted Answer

$\frac{dy}{dx}=-\frac{y}{6y+2x}$; normal $y=18x-107$

2021 RI Promo Q2

2021 RI Promo Q2

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