A curve $C$ has equation ${{\left( \ln y \right)}^{2}}-3xy+{{x}^{2}}-4=0$. Show that $\frac{\mathrm{

Question

A curve $C$ has equation ${{\left( \ln y \right)}^{2}}-3xy+{{x}^{2}}-4=0$. Show that $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\left( 3y-2x \right)y}{2\ln y-3xy}$. [3] It is given that $C$ contains the point $A$ with coordinates $\left( \alpha ,1 \right)$ where $\alpha Find the value of $\alpha $ and hence the equation of the normal to $C$ at the point $A$. [3]

Accepted Answer

(i) $\frac{dy}{dx}=\frac{(3y-2x)y}{2\ln y-3xy}$; (ii) $y=-\frac35x+\frac25$

2019 ASRJC Promo Q3

2019 ASRJC Promo Q3

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