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

Question

A curve $C$ has equation $\frac{{{x}^{2}}-4{{y}^{2}}}{{{x}^{2}}+x{{y}^{2}}}=\frac{1}{2}$. Show that $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{2x-{{y}^{2}}}{2xy+16y}$. [3] The points $P$ and $Q$ on $C$ each have $x$-coordinate $1$. The tangents to $C$ at $P$ and $Q$ meet at the point $N$. Find the exact coordinates of $N$. [6]

Accepted Answer

(i) dy/dx = (2x - y²)/(16y + 2xy); (ii) N = (-1/17, 0)

N2018 P1 Q7

N2018 P1 Q7

Video Solution:

Default solution