2011 H2 Mathematics Paper 1 Question 3

Differentiation I: Tangents and Normals, Parametric Curves

Answers

{y = -\frac{1}{p^3}x + \frac{3}{p}}

{Q \left( 3p^2 , 0 \right), }

{R \left( 0 , \frac{3}{p} \right)}

{x = \frac{27}{8y^2}}

\begin{align*} \frac{\mathrm{d}x}{\mathrm{d}t} &= 2 t \\ \frac{\mathrm{d}y}{\mathrm{d}t} &= - \frac{2}{t^{2}} \\ \end{align*}

\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} &= \frac{\mathrm{d}y}{\mathrm{d}t} \div \frac{\mathrm{d}x}{\mathrm{d}t} \\ &=- \frac{2}{t^{2}} \div 2 t \\ &=- \frac{1}{t^{3}} \\ \end{align*}

At point

{P \left( p^2, \frac{2}{p} \right), }

{t=p}

\frac{\mathrm{d}y}{\mathrm{d}x} = - \frac{1}{p^{3}}

Tangent at

{P\left( p^2, \frac{2}{p} \right):}

\begin{gather*} y - \frac{2}{p} = - \frac{1}{p^{3}} \left( x - p^2 \right) \\ y = -\frac{1}{p^3} x + \frac{1}{p} + \frac{2}{p} \\ \end{gather*}

Hence equation of tangent at

{P \left( p^2, \frac{2}{p} \right)}

y = -\frac{1}{p^3}x + \frac{3}{p} \; \blacksquare

At point

{Q, }

substituting

{y=0}

into equation of tangent at

{P,}

\begin{gather*} 0 = -\frac{1}{p^3}x + \frac{3}{p} \\ x = 3p^2 \end{gather*}

At point

{R, }

substituting

{x=0}

into equation of tangent at

{P,}

\begin{align*} y &= -\frac{1}{p^3}(0) + \frac{3}{p} \\ &= \frac{3}{p} \end{align*}

Hence coordinates of

{Q}

and

{R}

are:

\begin{align*} & Q \left( 3p^2 , 0 \right) \; \blacksquare \\ & R \left( 0 , \frac{3}{p} \right) \; \blacksquare \\ \end{align*}

Mid-point of

{QR:}

\left( \frac{3p^2}{2}, \frac{3}{2p} \right)

\begin{align} && \quad x &= \frac{3p^2}{2} \\ && \quad y &= \frac{3}{2p} \end{align}

From

{(2),}

\begin{equation} p = \frac{3}{2y} \end{equation}

Substituting

{(3)}

into

{(1),}

\begin{align*} x &= 3\left(\frac{3}{2y}\right)^2 \div 2 \\ &= \frac{27}{8y^2} \\ \end{align*}

Cartesian equation of the locus of the mid-point of

{QR}

{p}

varies:

x=\frac{27}{8y^2} \; \blacksquare