2018 H2 Mathematics Paper 1 Question 7

Differentiation I: Tangents and Normals, Parametric Curves

Answers

(ii)

N(−117,0){N \left( - \frac{1}{17}, 0 \right)}

Full solutions

(i)

x2−4y2x2+xy2=122x2−8y2=x2+xy2\begin{gather*} \frac{x^2-4y^2}{x^2+xy^2} = \frac{1}{2} \\ 2x^2 - 8y^2 = x^2 + xy^2 \\ \end{gather*}
x2−8y2=xy2\begin{equation} x^2 - 8y^2 = xy^2 \end{equation}
Differentiating implicitly w.r.t. x,{x,}
2x−16ydydx=y2+2xydydxdydx(2xy+16y)=2x−y2\begin{align*} 2x - 16y \frac{\mathrm{d}y}{\mathrm{d}x} = y^2 + 2xy\frac{\mathrm{d}y}{\mathrm{d}x} \\ \frac{\mathrm{d}y}{\mathrm{d}x} \left( 2xy+16y \right) = 2x-y^2 \end{align*}
dydx=2x−y22xy+16y  ■\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{2x-y^2}{2xy+16y}\;\blacksquare

(ii)

Substituting x=1{x=1} into (1),{(1),}
1−8y2=y29y2=1y=±13\begin{gather*} 1 - 8y^2 = y^2 \\ 9y^2 = 1 \\ y = \pm \frac{1}{3} \end{gather*}
When x=1,y=13,{x=1, y=\frac{1}{3},}
dydx=2(1)−192(1)(13)+16(13)=1754\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} &= \frac{2(1)-\frac{1}{9}}{2(1)\left(\frac{1}{3}\right)+16\left(\frac{1}{3}\right)} \\ &= \frac{17}{54} \end{align*}
When x=1,y=−13,{x=1, y=-\frac{1}{3},}
dydx=2(1)−192(1)(−13)+16(−13)=−1754\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} &= \frac{2(1)-\frac{1}{9}}{2(1)\left(-\frac{1}{3}\right)+16\left(-\frac{1}{3}\right)} \\ &= - \frac{17}{54} \end{align*}
Equations of tangents:
y−13=1754(x−1)y+13=−1754(x−1)\begin{align} && \quad y - \frac{1}{3} &= \frac{17}{54} \left( x - 1 \right) \\ && \quad y + \frac{1}{3} &= - \frac{17}{54} \left( x - 1 \right) \\ \end{align}
Solving with a GC, exact coordinates of N:{N:}
N(−117,0)  ■N \left( - \frac{1}{17}, 0 \right) \; \blacksquare