# Question 5:

$$\frac{x^2}{25}-\frac{y^2}{4}=1, \quad y=mx+c$$

## Part (a):

| Working/Answer | Mark | Notes |
|---|---|---|
| $\frac{x^2}{25}-\frac{(mx+c)^2}{4}=1 \Rightarrow 4x^2-25(m^2x^2+2cmx+c^2)=100$ | M1 | Substitutes to obtain a quadratic in $x$ and eliminates fractions |
| $(25m^2-4)x^2+50cmx+25c^2+100=0$ | A1 | Correct 3TQ |
| $"b^2=4ac" \Rightarrow (50cm)^2=4(25m^2-4)(25c^2+100)$ | M1 | Uses $b^2=4ac$ or equivalent |
| $2500c^2m^2 = 2500c^2m^2+10000m^2-400c^2-1600 \Rightarrow 10000m^2=400c^2+1600 \Rightarrow 25m^2=c^2+4$ | A1* | Fully correct proof with no errors |

### ALT 1 (Hyperbolic parameters):

| Working/Answer | Mark | Notes |
|---|---|---|
| $x=5\cosh t,\ y=2\sinh t \Rightarrow \frac{dy}{dx}=\frac{2\cosh t}{5\sinh t}$; tangent: $\frac{2\cosh t}{5\sinh t}(x-5\cosh t)=y-2\sinh t$ | M1A1 | M1: Attempts equation of tangent; A1: Correct equation (no simplification needed) |
| $25m^2 = \frac{4\cosh^2 t}{\sinh^2 t}$, $\ 4+c^2 = \frac{4\cosh^2 t}{\sinh^2 t}$ | M1 | Extracts $25m^2$ and $4+c^2$ from their equation |
| $\therefore 25m^2 = 4+c^2$ | A1* | Fully correct proof with no errors |

### ALT 2 (Trigonometric parameters):

| Working/Answer | Mark | Notes |
|---|---|---|
| $x=5\sec t,\ y=2\tan t \Rightarrow \frac{dy}{dx}=\frac{2\sec t}{5\tan t}$; tangent: $\frac{2\sec t}{5\tan t}(x-5\sec t)=y-2\tan t$ | M1A1 | M1: Attempts equation of tangent; A1: Correct equation (no simplification needed) |
| $25m^2=\frac{4\sec^2 t}{\tan^2 t}=\frac{4}{\sin^2 t}$, $\ 4+c^2=\frac{4}{\sin^2 t}$ | M1 | Extracts $25m^2$ and $4+c^2$ from their equation |
| $\therefore 25m^2 = 4+c^2$ | A1* | Fully correct proof with no errors |

# Question 5(b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $25m^2 = c^2 + 4$ and $2 = m + c$, giving $25m^2 = (2-m)^2 + 4$ or $25(2-c)^2 = c^2 + 4$ | M1 | Uses the given hyperbola and straight line with result from (a) to obtain equation in $m$ or $c$ |
| $24m^2 + 4m - 8 = 0$ or $24c^2 - 100c + 96 = 0$ | A1 | Correct 3TQ in $m$ or $c$ |
| $24m^2 + 4m - 8 = 0 \Rightarrow m = \frac{1}{2}, -\frac{2}{3}$ or $24c^2 - 100c + 96 = 0 \Rightarrow c = \frac{3}{2}, \frac{8}{3}$ | M1 | Solves their 3TQ in $m$ or $c$ |
| $y = \frac{1}{2}x + \frac{3}{2}$ **or** $y = -\frac{2}{3}x + \frac{8}{3}$ | A1 | One correct tangent |
| $y = \frac{1}{2}x + \frac{3}{2}$ **and** $y = -\frac{2}{3}x + \frac{8}{3}$ | A1 | Both correct tangents |

**Total: (5)**

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# Question 5(c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $m = \frac{1}{2}, c = \frac{3}{2} \Rightarrow \frac{9}{4}x^2 + \frac{75}{2}x + \frac{625}{4} = 0 \Rightarrow x = \ldots$ or $m = -\frac{2}{3}, c = \frac{8}{3} \Rightarrow \frac{64}{9}x^2 - \frac{800}{9}x + \frac{2500}{9} = 0 \Rightarrow x = \ldots$ | M1 | Uses one of their $m$ and $c$ pairs and solves for $x$ |
| $x = -\frac{25}{3}, y = -\frac{8}{3}$ **or** $x = \frac{25}{4}, y = -\frac{3}{2}$ | A1 | One correct point |
| $x = -\frac{25}{3}, y = -\frac{8}{3}$ **and** $x = \frac{25}{4}, y = -\frac{3}{2}$ | A1 | Both correct points |

**Total: (3) — Question Total: 12**

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