AS June 2023 Q6
6. The parabola \(C\) has equation \(y^2 = 4ax\) where \(a\) is a positive constant.
The point \(P(at^2, 2at)\), \(t \neq 0\), lies on \(C\)
The normal to \(C\) at \(P\) is parallel to the line with equation \(y = 2x\)
The normal to \(C\) at \(P\) intersects \(C\) again when \(x = 9\)
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{2a}{2at}\) or \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 2\sqrt{a} \times \dfrac{1}{2}x^{-\frac{1}{2}}\) or \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{2a}{y}\) | B1 | 1.1b |
| Finds the perpendicular gradient at sets equal to 2 to find a value of \(t\) \(-t = 2 \Rightarrow t = -2\ *\) | M1 A1* | 2.1 1.1b |
| (3) |
Notes
B1: any correct expression for \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\), may be unsimplified.
M1: Finds the perpendicular gradient and sets equal to 2 and uses \((at^2, 2at)\) to find a value of \(t\).
Alternatively set the gradient equal to – ½ to find a value of \(t\). May be seen as part of a longer method finding the equation of the tangent first.
A1*cso: Arrives at correct value \(t = -2\) only from correct work , no errors seen.
Alt to (a)
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{2a}{2at}\) or \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 2\sqrt{a} \times \dfrac{1}{2}x^{-\frac{1}{2}}\) or \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{2a}{y}\) | B1 | 1.1b |
| \(t = -2 \Rightarrow y = -4a \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{2a}{-4a} = -\dfrac{1}{2}\) so gradient of normal is \(-\dfrac{1}{-\frac{1}{2}} = 2\), hence normal parallel to \(y = 2x\) (when \(t = -2\))* | M1 A1* | 2.1 1.1b |
| (3) |
B1: As main scheme.
M1: Finds the value of the normal when \(t = -2\)
A1: Correct work to show normal has gradient 2 when \(t = -2\) and a conclusion.
| Scheme | Marks | AO |
|---|---|---|
| \(t = -2 \Rightarrow (4a, -4a)\) Correct equation of the normal \(y + 4a = 2(x - 4a)\) | B1 | 1.1b |
| \(2x - 12a = \sqrt{4ax} \Rightarrow 2(9) - 12a = \sqrt{4a(9)} \Rightarrow 18 - 12a = 6\sqrt{a}\) \(\Rightarrow 12a + 6\sqrt{a} - 18 = 0\) or \((2x - 12a)^2 = 4ax \Rightarrow (2(9) - 12a)^2 = 4a(9) \Rightarrow (18 - 12a)^2 = 36a\) \(\Rightarrow 144a^2 - 468a + 324 = 0\) | M1 | 3.1a |
| Solves 3TQ to find a value for \(a\) Note: Quadratic for \(\sqrt{a}\) leads to \(\sqrt{a} = 1\) and \(-1.5\) Quadratic for \(a\) leads to \(a = 1\) and \(\dfrac{9}{4}\) | dM1 A1 | 1.1b 1.1b |
| e.g. for \(t = -2, a = \dfrac{9}{4}\) leads to \(x = 9\) but \(P\) is the other intersection point (\(a = 1\) leads to the correct coordinate of \(P(4, -4)\), or \(\sqrt{a}\) cannot be negaitive) therefore \(a = 1\) | A1 | 2.4 |
| (5) | ||
| (8 marks) |
Notes
B1: Correct equation of the normal. May be implied by later working.
M1: Starts the process of solving simultaneously the equations of the parabola and normal by eliminating \(y\), substituting \(x = 9\) and forming 3TQ equation for \(\sqrt{a}\) or \(a\). There may be slips in the process.
dM1: Dependent on previous mark. Solves their 3TQ to find a value for \(a\)
A1: Correct possible values for \(a\)
A1: Deduces the correct value of \(a\) and gives a correct reason.
NB: Use of gradient \(-2\) instead of 2 in (b) often leads to \(a = 9\), and can score maximum B0M1M1A0A0.
Alt 1 to (b)
| Scheme | Marks | AO |
|---|---|---|
| \(t = -2 \Rightarrow (4a, -4a)\) Correct equation of the normal \(y + 4a = 2(x - 4a)\) | B1 | 1.1b |
| \((2x - 12a)^2 = 4ax \Rightarrow 4x^2 - 48ax + 144a^2 = 4ax\) \(\Rightarrow x^2 - 13ax + 36a^2 = 0\) | M1 | 3.1a |
| \(\Rightarrow (x - 4a)(x - 9a) = 0 \Rightarrow x = 4a, 9a\) | dM1 A1 | 1.1b 1.1b |
| \(x = 4a\) is \(P\) so other point is \(x = 9a\) so \(9 = 9a \Rightarrow a = 1\) | A1 | 2.4 |
| (5) |
B1: Correct equation of the normal.
M1: Starts the process of solving simultaneously the equations of the parabola and normal by eliminating \(y\), and expanding to a quadratic in \(a\) and \(x\).
dM1: Dependent on previous mark. Solves their 3TQ in \(x\) to find the possible values of \(x\) in terms of \(a\)
A1: Correct values of \(x\) in terms of \(a\)
A1: Deduces the correct value of \(a\) with a minimal reason or clear correct working shown.
Alt 2 to (b)
| Scheme | Marks | AO |
|---|---|---|
| \(t = -2 \Rightarrow (4a, -4a)\) Correct equation of the normal \(y + 4a = 2(x - 4a)\) | B1 | 1.1b |
| \(2at + 4a = 2(at^2 - 4a) \Rightarrow 2t + 4 = 2t^2 - 8\) \(\Rightarrow 2t^2 - 2t - 12 = 0\) | M1 | 3.1a |
| \(\Rightarrow t^2 - t - 6 = 0 \Rightarrow (t + 2)(t - 3) = 0 \Rightarrow t = \text{“}3\text{”}\) | dM1 A1 | 1.1b 1.1b |
| \(t = -2\) is \(P\) so other point is when \(t = 3\) therefore \(9 = a \times 3^2 \Rightarrow a = 1\) | A1 | 2.4 |
| (5) |
B1: Correct equation of the normal.
M1: Substitutes the parametric equations for the parabola into the equation for the normal and cancels the factors \(a\) to produce an equation in \(t\) only.
dM1: Dependent on previous mark. Solves their 3TQ in \(t\) to find at least the other value for \(t\)
A1: Correct other value for \(t\)
A1: Deduces the correct value of \(a\) with a minimal reason for choice of \(t\) or clear correct working shown.
(Brackets that are missing from the printed mark scheme because of a font fault have been restored.)