FP3 June 2018 Q7
7. The ellipse \(E\) has foci at the points \((\pm 3, 0)\) and has directrices with equations \(x = \pm\dfrac{25}{3}\)
The straight line \(l\) has equation \(y = mx + c\), where \(m\) and \(c\) are positive constants.
Given that the line \(l\) is a tangent to \(E\),
The line \(l\) intersects the \(x\)-axis at the point \(A\) and intersects the \(y\)-axis at the point \(B\).
| Scheme | Marks |
|---|---|
| \(ae = 3,\ \dfrac{a}{e} = \dfrac{25}{3}\) or \(ae = \pm 3,\ \dfrac{a}{e} = \pm\dfrac{25}{3}\) | B1 |
| \(e^2 = \dfrac{9}{25}\) and \(a^2 = 25\) | M1 |
| \(b^2 = a^2\left(1 - e^2\right) \Rightarrow b^2 = 25\left(1 - \dfrac{9}{25}\right) = 16\) | M1 |
| \(\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1 \Rightarrow \dfrac{x^2}{25} + \dfrac{y^2}{16} = 1\ \left(\text{or } \dfrac{x^2}{5^2} + \dfrac{y^2}{4^2} = 1\right)\) | M1A1 |
| (5) |
Notes
B1: Correct equations. (Ignore the use of + or − throughout)
M1: Solves to find \(a\) or \(a^2\) and \(e\) or \(e^2\)
M1: Uses correct eccentricity formula to find \(b\) or \(b^2\)
M1: Uses a correct ellipse formula and their \(a\) and \(b\)
A1: Correct equation
| Scheme | Marks |
|---|---|
| \(\dfrac{x^2}{25} + \dfrac{(mx + c)^2}{16} = 1\) | M1 |
| \(16x^2 + 25\left(m^2x^2 + 2mcx + c^2\right) = 400\) \(\therefore \left(16 + 25m^2\right)x^2 + 50mcx + 25\left(c^2 - 16\right) = 0\) * | A1* |
| (2) |
Notes
M1: Substitutes for \(y\). Allow in terms of \(a\) and \(b\).
A1*: Correct proof including sufficient intermediate working (at least one step) with no errors.
| Scheme | Marks |
|---|---|
| \(b^2 - 4ac = 0 \Rightarrow (50mc)^2 - 4\left(16 + 25m^2\right)\left(25\left(c^2 - 16\right)\right) = 0\) | M1A1 |
| \(c^2 = 25m^2 + 16\) | A1 |
| (3) |
Notes
M1: Uses \(b^2 - 4ac = 0\) oe e.g. \(b^2 = 4ac\) with the given quadratic (may be implied by their equation)
Do not allow as part of an attempt to use the quadratic formula unless the discriminant is “extracted” and used = 0
A1: Correct equation (the “= 0” may be implied/appear later)
The equation above scores M1A1
A1: cao
| Scheme | Marks |
|---|---|
| \(x = \pm\dfrac{\sqrt{25m^2 + 16}}{m},\ y = \sqrt{25m^2 + 16}\) | B1ft |
| Area \(OAB\ (= T) = \dfrac{1}{2}\dfrac{\sqrt{25m^2 + 16}}{m}\sqrt{25m^2 + 16}\) | M1 |
| \(T = \dfrac{25m^2 + 16}{2m}\) * | A1* |
| (3) |
Notes
B1ft: Follow through their \(p\) and \(q\). May be implied by their attempt at the triangle area.
M1: Correct triangle area method (Allow \(\pm\) area here)
A1*: Correct area. (Must be positive)
(d) Alt 1
| Scheme | Marks |
|---|---|
| \(y = mx + c \Rightarrow y = c,\ x = \pm\dfrac{c}{m}\) | B1 |
| Area \(OAB\ (= T) = \dfrac{1}{2} \times c \times \dfrac{c}{m} = \dfrac{c^2}{2m}\) | M1 |
| \(T = \dfrac{25m^2 + 16}{2m}\) * | A1* |
| (3) |
B1: Correct intercepts
M1: Correct triangle area method (Allow \(\pm\) area here)
A1*: Correct positive area. Must follow the final A1 in part (c) unless the work for part (c) is done in part (d).
| Scheme | Marks |
|---|---|
| \(\dfrac{\mathrm{d}T}{\mathrm{d}m} = \dfrac{25}{2} - \dfrac{8}{m^2} = 0 \Rightarrow m = \dfrac{4}{5}\) or \(\dfrac{\mathrm{d}T}{\mathrm{d}m} = \dfrac{2m(50m) - 2\left(25m^2 + 16\right)}{4m^2} = 0 \Rightarrow m = \dfrac{4}{5}\) | M1 |
| \(m = \dfrac{4}{5} \Rightarrow T = 20\) | A1 |
| (2) | |
| (15 marks) |
Notes
M1: Solves \(\dfrac{\mathrm{d}T}{\mathrm{d}m} = 0\) to obtain a value for \(m\)
A1: cao
Alternative for (e)
| Scheme | Marks |
|---|---|
| \(T = \dfrac{25m^2 + 16}{2m} = \dfrac{(5m - 4)^2 + 40m}{2m},\ (5m - 4)^2 = 0 \Rightarrow T = \dfrac{40m}{2m}\) | M1 |
| \(T = 20\) | A1 |
| (2) |
M1: Writes \(T\) as \(\dfrac{(5m - 4)^2 + \ldots}{2m}\) and realises minimum when \((5m - 4) = 0\)
A1: cao