(a) $y = \frac{25}{x}$, so $\frac{dy}{dx} = -25x^{-2}$ | M1 |

$\frac{dy}{dx} = -\frac{25}{(5p)^2} = -\frac{1}{p^2}$ | A1 |

$y - \frac{5}{p} = -\frac{1}{p^2}(x - 5p) \Rightarrow p^2y + x = 10p$ | M1 A1 | (*) (4)

(b) $q^2y + x = 10q$ only | B1 | (1)

(c) $(p^2 - q^2)y = 10(p-q)$, so $y = \frac{10(p-q)}{(p^2-q^2)} = \frac{10}{p+q}$ | M1 A1cso |

$x = 10p - p^2 \cdot \frac{10}{p+q} = \frac{10pq}{p+q}$ | M1 A1 cso | (4)

(d) Line $PQ$ has gradient $\frac{5}{p} - \frac{5}{q} = -\frac{1}{pq}$ | M1 A1 |

$ON$ has gradient $\frac{10}{p+q} \cdot \frac{1}{10pq} = \frac{1}{pq}$ or $\frac{-1}{(-1)} = pq$ could be unsimplified | B1 |

equivalents seen anywhere

As these lines are perpendicular $-\frac{1}{pq} \times -\frac{1}{pq} = -1$, so $p^2q^2 = 1$ | M1 A1 |

**OR for ON:**
$y - y_1 = m(x - x_1)$ with gradient (equivalent to) $pq$ and sub in points $O$ AND $N$ to give $p^2q^2 = 1$

**OR for PQ:**
$y - y_1 = m(x - x_1)$ with gradient (equivalent to) $-pq$ and sub in points $P$ and $Q$ to give $p^2q^2 = 1$. NB -$pq$ used as gradient of $PQ$ implies first M1A1 | M1 A1 |

| (5) [14] |

**Alternatives for first M1 A1 in part (a):**

$x\frac{dy}{dx} + y = 0 \Rightarrow \frac{dy}{dx} = -\frac{y}{x}$ | M1 |

So at $P$ gradient $= -\frac{5p}{5} = -\frac{1}{p^2}$ | A1 |

Or $x = 5t, y = \frac{5}{t} \Rightarrow \frac{dx}{dt} = 5, \frac{dy}{dt} = -\frac{5}{t^2}$ so $\frac{dy}{dx} = -\frac{5}{t^2} \cdot \frac{1}{5} = -\frac{1}{t^2}$ | M1 A1 |

**Notes:**

(a) First M for attempt at explicit, implicit or parametric differentiation not using $p$ or $q$ as an initial parameter; first A for $-\frac{1}{p^2}$ or equivalent. Quoting gradient award first M0A0. Second M for using $y - y_1 = m(x - x_1)$ and attempt to substitute or $y = mx + c$ and attempt to find $c$; gradient in terms of $p$ only and using $(5p, \frac{5}{p})$; second A for correct solution only.

(c) First M for eliminating $x$ and reaching $y = f(p,q)$; second M for eliminating $y$ and reaching $x = f(p,q)$; both As for given answers. Minimum amount of working given in the main scheme above for 4/4, but do not award accuracy if any errors are made.

(d) First M for use of $\frac{y_2 - y_1}{x_2 - x_1}$ and substituting; first A for $-\frac{1}{pq}$ or unsimplified equivalent.

Second M for their product of gradients=-1 (or equating equivalent gradients of $ON$ or equating equivalent gradients of $PQ$); second A for correct answer only.

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