# Question 1:

## Part (a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $ae = \frac{13}{2}$ **or** $\frac{a}{e} = \frac{72}{13}$ | B1 | One correct equation in $a$ and $e$. Allow equivalent correct equations. Could include $-$ or $\pm$ signs |
| e.g., $a = \frac{72}{13}e \Rightarrow \frac{72}{13}e^2 = \frac{13}{2} \Rightarrow e^2 = \ldots \left(\frac{169}{144}\right)$ or $a = \frac{13}{2e} \Rightarrow \frac{13}{2e^2} = \frac{72}{13} \Rightarrow e^2 = \ldots \left(\frac{169}{144}\right)$ | M1 | Having obtained two equations in $a$ and $e$ of correct form i.e., $ae = p$ and $\frac{a}{e} = q$, $p, q \neq 0$, solves simultaneously to find a positive value for $e^2$ (no requirement for $e > 1$) or $e$. Condone poor algebra provided a value is obtained. May find $a$ first. |
| $e = \frac{13}{12}$ or $1\frac{1}{12}$ or $1.08\overline{3}$. Not $\pm\frac{13}{12}$ unless negative value clearly rejected. | A1 | |

## Part (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\left\{a = \frac{72}{13} \times \frac{13}{12} = 6\right\}$ or $\left\{a = \frac{13}{2\left(\frac{13}{12}\right)} = 6\right\}$; $b^2 = a^2(e^2-1) = \ldots$; $\left\{b^2 = 6^2\left(\left(\frac{13}{12}\right)^2 - 1\right) = \frac{25}{4}\right\}$ or $b = \frac{5}{2}$ | M1 | With any value for $a$, which might be seen in part (a), and their $e$, **uses** a correct eccentricity formula with correct substitution to find a value for $b^2$ or $b$. Could be implied. May see $b = a\sqrt{e^2-1}$ or use of $e = \sqrt{1+\frac{b^2}{a^2}}$ or $e = \frac{c}{a}$ with $c = \sqrt{a^2+b^2}$ |
| $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \Rightarrow \frac{x^2}{36} - \frac{y^2}{\frac{25}{4}} = 1$ | M1 | Applies $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ **correctly** for their values. Not dependent. Could use $b^2x^2 - a^2y^2 = a^2b^2$ |
| e.g., $25x^2 - 144y^2 = 900$ | A1 | A correct **equation** in correct form. Requires all previous 5 marks but allow if 4 marks with A0 in (a) for $e = \pm\frac{13}{12}$ and negative value not rejected in part (a). Any positive integer multiple. Allow equivalents provided variables on one side and constant on the other and $y^2$ term has negative coefficient. Just $p=25, q=144, r=900$ requires $px^2 - qy^2 = r$ to be seen. Ignore wrong values for $p, q, r$ following a correct equation (e.g., "$q = -144$") |

**Alt using $PS^2 = e^2PM^2$:**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\left(x - \frac{13}{2}\right)^2 + y^2 = \left(\frac{13}{12}\right)^2\left(x - \frac{72}{13}\right)^2$ | M1 | Forms equation correct for their $ae$, $e$ and $\frac{a}{e}$ |
| $x^2 - 13x + \frac{169}{4} + y^2 = \frac{169}{144}x^2 - 13x + 36 \Rightarrow \frac{25}{144}x^2 - y^2 = \frac{25}{4}$ | M1 | Expands and reaches $rx^2 - sy^2 = t$, $r, s, t \neq 0$ |
| e.g., $25x^2 - 144y^2 = 900$ | A1 | as main scheme |

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