2018 H2 Maths Paper 1 Q7 — Worked Solution

Part (i)

3 marks

Approach

The equation for $C$ has $y$ terms buried inside a fraction, which is awkward to differentiate directly. We first clear denominators to get a polynomial relation between $x$ and $y$ , then differentiate implicitly.

Since $\frac{x^{2} - 4y^{2}}{x^{2} + xy^{2}} = \frac{1}{2}$ , cross-multiplying,

2\left( x^{2} - 4y^{2} \right) = x^{2} + xy^{2}

2x^{2} - 8y^{2} = x^{2} + xy^{2}

x^{2} - 8y^{2} - xy^{2} = 0

x^{2} = y^{2}(x + 8).

Differentiating both sides with respect to $x$ , using the product rule on the right (treating $y$ as a function of $x$ ),

2x = 2y\frac{dy}{dx}(x + 8) + y^{2}

2y(x + 8)\frac{dy}{dx} = 2x - y^{2}

\frac{dy}{dx} = \frac{2x - y^{2}}{2y(x + 8)}.

Since $2y(x + 8) = 2xy + 16y$ ,

\frac{dy}{dx} = \frac{2x - y^{2}}{2xy + 16y}.\ \ \text{(shown)}

Part (ii)

6 marks

Approach

With $x = 1$ fixed, the relation $x^{2} = y^{2}(x + 8)$ from (i) gives the two $y$ -values of $P$ and $Q$ directly. We then find the gradient at each point from the given derivative, write down both tangents, and solve them simultaneously for $N$ .

Since $x = 1$ lies on $C$ , substituting into $x^{2} = y^{2}(x + 8)$ ,

1 = y^{2}(9)

y^{2} = \frac{1}{9}

y = \pm \frac{1}{3}.

Take $P = \left( 1,\ \frac{1}{3} \right)$ and $Q = \left( 1,\ - \frac{1}{3} \right)$ .

At $P$ , using the result of (i) with $x = 1$ , $y = \frac{1}{3}$ ,

\left. \ \frac{dy}{dx} \right|_{P} = \frac{2(1) - \left( \frac{1}{3} \right)^{2}}{2(1)\left( \frac{1}{3} \right) + 16\left( \frac{1}{3} \right)} = \frac{\frac{17}{9}}{6} = \frac{17}{54}.

At $Q$ , with $x = 1$ , $y = - \frac{1}{3}$ , the numerator is unchanged (it depends only on $y^{2}$ ) while the denominator changes sign, so

\left. \ \frac{dy}{dx} \right|_{Q} = - \frac{17}{54}.

The tangent to $C$ at $P$ is

y - \frac{1}{3} = \frac{17}{54}(x - 1),

and the tangent to $C$ at $Q$ is

y + \frac{1}{3} = - \frac{17}{54}(x - 1).

Note: Since these two tangents have equal and opposite gradients and $P$ , $Q$ are reflections of each other in the $x$ -axis, adding the two equations eliminates the $x$ -terms and shows $N$ lies on the $x$ -axis; we solve directly to confirm this and find the $x$ -coordinate.

Solving the two tangent equations simultaneously: from the first, $y = \frac{1}{3} + \frac{17}{54}(x - 1)$ ; substituting into the second,

\frac{1}{3} + \frac{17}{54}(x - 1) + \frac{1}{3} = - \frac{17}{54}(x - 1)

\frac{34}{54}(x - 1) = - \frac{2}{3}

x - 1 = - \frac{2}{3} \cdot \frac{54}{34} = - \frac{18}{17}

x = - \frac{1}{17}.

Substituting back,

y = \frac{1}{3} + \frac{17}{54}\left( - \frac{18}{17} \right) = \frac{1}{3} - \frac{1}{3} = 0.

To check your answer:

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

lies on the

x

-axis, as expected from the symmetry of

C

(unchanged under

y \mapsto - y

) between the mirror-image points

P

and

Q

. ✓

$\therefore N = \left( - \frac{1}{17},\ 0 \right)$ .

2018 H2 Maths Paper 1, Question 7

Part (i)

Part (ii)

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