2014 H2 Mathematics Paper 1 Question 2

Differentiation I: Tangents and Normals, Parametric Curves

Answers

{\left(-3, -3 \right)}

\begin{equation} \qquad x^2y + xy^2 + 54 = 0 \end{equation}

Differentiating implicitly w.r.t.

{x,}

x^2 \frac{\mathrm{d}y}{\mathrm{d}x} + 2xy + 2xy \frac{\mathrm{d}y}{\mathrm{d}x} + y^2 = 0

When the gradient is

{-1,}

\begin{gather*} x^2(-1)+2xy+2xy(-1)+y^2 = 0 \\ -x^2 + y^2 = 0 \\ y^2 = x^2 \end{gather*}

\begin{align} && \quad y &= x \quad \textrm{ or} \\ && \quad y &= -x \end{align}

Substituting

{(2)}

into

{(1),}

\begin{align*} x^3 + x^3 + 54 &= 0 \\ 2x^3 &= -54 \\ x^3 &= -27 \\ x &= -3 \\ y &= -3 \end{align*}

Substituting

{(3)}

into

{(1),}

\begin{align*} \textrm{LHS} &= x^2(-x) + x(-x)^2 + 54 \\ &= -x^3 + x^3 + 54 \\ &= 54 \\ &\neq 0 = \textrm{RHS} \end{align*}

Hence the only solution to the system of equations is

{x=y=-3, }

so there is only on such point such that the gradient is

{-1 \; \blacksquare}

Coordinates of point on

{C}

at which the gradient is

{-1: }

\left(-3, -3 \right) \; \blacksquare