A curve $C$ has equation $3x^2 + 2xy - 4y^2 = 1$. Show that \(\frac{\mathrm{d}y}{\mathrm{d}

Question

A curve $C$ has equation $3x^2 + 2xy - 4y^2 = 1$. Show that $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{3x + y}{4y - x}$.[2] Find the acute angle between the tangent to $C$ at the point $(1, 1)$ and the $x$-axis.[2] Show that there is no point on $C$ where the tangent to the curve is parallel to the $x$-axis.[3]

Accepted Answer

(a) $\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{3x+y}{4y-x}$ (b) $53.1^\circ$ (c) No such point exists

2024 VJC Promo Q5

2024 VJC Promo Q5

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