A curve $C$ has equation ${{x}^{2}}-xy+{{y}^{2}}-\alpha =0$, where $\alpha $ is a positive real numb

Question

A curve $C$ has equation ${{x}^{2}}-xy+{{y}^{2}}-\alpha =0$, where $\alpha $ is a positive real number. Show that $\left( 2y-x \right)\frac{\mathrm{d}y}{\mathrm{d}x}=y-2x$. [1] It is given that $x=2\sqrt{3}$ is a tangent to the curve $C$. Show that $\alpha =9$. [2] With this value of $\alpha $, Find the exact coordinates of the stationary points. [3] By further differentiation of the equation in (i), determine the nature of the stationary points found in (iii). [3]

Accepted Answer

(i) Shown (ii) a = 9 (iii) $(\sqrt{3}, 2\sqrt{3})$ max, $(-\sqrt{3}, -2\sqrt{3})$ min

2018 AJC Promo Q5

2018 AJC Promo Q5

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