## Question 12(a):

| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{dx}{dt} = \frac{-1}{t^2}$, $\frac{dy}{dt} = 2$; $\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$ | M1 | 1.1a — Attempt correct process to find gradient in terms of $t$ or $p$. Need $\frac{dx}{dt} = kt^{-2}$ and $\frac{dy}{dt} = 2$. **SC B1** for gradient of $-2x^2$ if never seen in terms of $t$ or $p$. |
| $\frac{dy}{dx} = -2t^2$ | A1 | 2.1 — Obtain correct gradient in terms of $t$ or $p$. |
| $y - 2p = -2p^2\left(x - \frac{1}{p}\right)$ | M1 | 1.1a — Attempt equation of tangent. Condone still working in terms of $t$. Allow mixture of $t$ and $p$ as long as convincingly recovered. Using their gradient from a differentiation attempt, but not dependent on first M1. |
| $y = -2p^2x + 4p$ **A.G.** | A1 | 2.1 — Obtain given answer. Expand brackets and simplify to given answer, or find $c$ and substitute back into equation. |

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## Question 12(b):

| Answer | Mark | Guidance |
|--------|------|----------|
| $m' = \frac{1}{2p^2}$ | B1FT | 1.1a — Correct (unsimplified) gradient of normal, following their derivative. Gradient in terms of $t$ or $p$, but not $x$. Could FT on their incorrect derivative or deduce from the equation given in (a). |
| $y - 2p = \frac{1}{2p^2}\left(x - \frac{1}{p}\right)$; $y = \frac{1}{2p^3}x + 2p - \frac{1}{2p^3}$ | M1 | 1.1 — Attempt equation of normal. Allow mixture of $t$ and $p$ as long as convincingly recovered. Substitution into $y - y_1 = m(x - x_1)$ or equation involving $c$ from $y = mx + c$. |
| At $B$, $y = 0$ so $x = 2p^2\!\left(\frac{1}{2p^3} - 2p\right) = \frac{1}{p} - 4p^3$ | M1, A1 | 3.1a, 2.1 — Use $y = 0$ to attempt $x$-coordinate of $B$; correct $x$-coordinate for $B$. Any equivalent form. |
| At $A$, $y = 0$ so $x = \frac{4p}{2p^2} = \frac{2}{p}$ | B1 | 2.1 — Correct $x$-coordinate for $A$. Any equivalent form. |
| $PA = \sqrt{\left(\frac{1}{p}\right)^2 + (2p)^2}$; $PB = \sqrt{(4p^3)^2 + (2p)^2}$ | M1 | 3.1a — Attempt length of $PA$ or $PB$. Must use correct distance formula, using given $P$ and their coordinates for $A$ and/or $B$, which must involve a function of $p$. |
| $PA$ and $PB$ correct | A1 | 2.1 — Or correct $(PA)^2$ and $(PB)^2$. |
| $PA : PB = \frac{1}{p}\sqrt{4p^4+1} : 2p\sqrt{4p^4+1} = \frac{1}{p} : 2p = 1 : 2p^2$ **A.G.** | A1 | 2.1 — Simplify ratio to obtain given answer. Must show clear method, such as same expression in each square root before cancelling. Could also consider fraction then cancel to deduce given ratio. Could simplify $(PA)^2 : (PB)^2$, then square root to obtain ratio. |