# Question 9:

## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y=\frac{25}{x} \Rightarrow \frac{dy}{dx}=-25x^{-2}$, or $y+x\frac{dy}{dx}=0 \Rightarrow \frac{dy}{dx}=-\frac{y}{x}$, or $\dot{x}=5,\ \dot{y}=-\frac{5}{t^2}$ so $\frac{dy}{dx}=-\frac{1}{t^2}$ | B1 | Any correct expression for gradient of tangent |
| At $P$: $\frac{dy}{dx}=-\frac{1}{p^2}$ so gradient of normal is $p^2$ | M1, A1 | M1: Substitutes into derived expression using calculus for gradient of normal at $P$; A1: cao |
| Either $y-\frac{5}{p}=p^2(x-5p)$ or $y=p^2x+k$ using $x=5p,\ y=\frac{5}{p}$ | M1 | Use of formula for straight line with changed gradient |
| $y - p^2x = \frac{5}{p}-5p^3$ | A1 cso | |

## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| At point $A$: $y+p^2y=\frac{5}{p}-5p^3$ or $-x-p^2x=\frac{5}{p}-5p^3$ | M1 | Replaces $x$ by $-y$ or $y$ by $-x$ |
| $y(1+p^2)=\frac{5}{p}(1-p^4)$ or $-x(1+p^2)=\frac{5}{p}(1-p^4)$ | M1 | Factorises $(1-p^4)$ to simplify |
| $y=\frac{5}{p}(1-p^2)=\frac{5}{p}-5p$ and $x=-\frac{5}{p}(1-p^2)=-\frac{5}{p}+5p$ | A1 cso | Obtains both $x$ and $y$; no errors |

## Part (c)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $M$ has coordinates $\left(-\frac{5}{2p}+5p,\ \frac{5}{p}-\frac{5p}{2}\right)$ o.e. | B1 | Correct $x$-coordinate and correct $y$-coordinate of midpoint |
| When $y=0$: $\frac{5}{p}-\frac{5p}{2}=0$ giving $p=\sqrt{2}$, so $M$ has $x$-coordinate $\frac{15}{4}\sqrt{2}$ o.e. | M1, A1 | M1: Sets $y=0$ and finds $p$; A1: $+\frac{15}{4}\sqrt{2}$ only |

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