A curve $C$ has equation $3{{x}^{2}}+2xy-{{y}^{2}}=80.$ Show that $\frac{\mathrm{d}y}{\mathrm{d}x}=\

Question

A curve $C$ has equation $3{{x}^{2}}+2xy-{{y}^{2}}=80.$ Show that $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{3x+y}{y-x}.$ [2] Show that the curve $C$ has no stationary points. [3] The normal to the curve at the point $P(6,-2)$ meets the curve again at the point $Q$. Find the coordinates of $Q$. [4]

Accepted Answer

(i) $\frac{\mathrm dy}{\mathrm dx}=\frac{3x+y}{y-x}$ (ii) No stationary points (iii) $Q=\left(-\frac{14}{3},-\frac{22}{3}\right)$

2015 DHS P1 Q8

2015 DHS P1 Q8

Video Solution:

Solution:

Solution locked