Notes

M1: Attempts to solve simultaneously by substituting \(y = mx + c\) into \(\dfrac{x^2}{64} + \dfrac{y^2}{36} = 1\) and attempts to multiply out \((mx + c)^2\) to obtain an equation in \(x\), \(m\) and \(c\)

A1: Proceeds to obtain the equation in the required form with the correct value of \(k\)

Notes

M1: Realises that for \(l\) to be a tangent, the discriminant must be zero and applies this to their equation correctly, may contain \(k\)

M1: Uses the other condition that \(l\) passes through \((10, 20)\) to eliminate \(c\) or \(m\) to obtain an equation in \(c\) or \(m\) only following an attempt at setting their discriminant \(= 0\)

A1: Correct 3TQ in \(c\) or \(m\)

dM1: Dependent on previous M. For a complete method to obtain at least one set of values for \(c\) and \(m\). I.e. Finds values for \(m\) or \(c\) from a 3TQ (no need to check the roots) and attempts to find the value of the other constant

A1: Both correct equations in any form

Notes

B1: Correct derivative in terms of \(t\) from a correct calculus method. Just stating \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{t}\) with no supporting working is B0.

M1: Uses \(y - 2at = \text{their ‘}\dfrac{1}{t}\text{’}\left(x - at^2\right)\), or uses \(y = \text{‘}\dfrac{1}{t}\text{’}x + c\) with \(\left(at^2, 2at\right)\) to find \(c\). They may have just stated the \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{t}\) for this mark.

A1*: Achieves the printed equation with no errors, but allow if the \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{t}\) is stated without working (so B0M1A1 is possible).

Notes

B1: Deduces a correct equation for one of the tangents. May find both but accept for one correct. If working in terms of \(a\) or \(a\) and \(t\) they must identify \(a = 2\) or \(at^2 = -2k^2\) (oe as relevant) at some stage to gain this mark following a correct tangent in terms of \(a\) or \(a\) and \(t\).

M1: Finds the intersection of the tangent with the \(x\)-axis, or solves both tangent equations simultaneously to find at least the \(x\) coordinate of the point of intersection. May be in terms of \(a\) and \(t\) for this mark.
Note the first 2 marks can be implied from stating a correct point of intersection

M1: A complete method to find the value of \(k\). E.g. uses \(\dfrac{8k \times \left(2k^2 - \text{their ‘}-2k^2\text{’}\right)}{2} = 432\) and solves to find a value for \(k\). Alternative approaches are possible, such as
\(432 = \dfrac{1}{2}\left|\overrightarrow{DC} \times \overrightarrow{DB}\right| = \dfrac{1}{2}\left\|\begin{matrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\ 4k^2 & 4k & 0\\ 4k^2 & -4k & 0\end{matrix}\right\| = \dfrac{1}{2}\left|-32k^3\mathbf{k}\right| = 16k^3 \Rightarrow k = \ldots\) or
\(432 = \dfrac{1}{2}\begin{vmatrix}-2k & 2k & 2k & -2k\\ 0 & 4k^2 & -4k^2 & 0\end{vmatrix} = \dfrac{1}{2}\left|\left(-8k^3 - 8k^3 + 0 - 0 - 8k^3 - 8k^3\right)\right| = 16k^3 \Rightarrow k = \ldots\)

A1: \(k = 3\)

A1 Deduces the correct coordinates \((18, 12)\) and \((18, -12)\) Do not be concerned with the labelling.

Notes

M1: A complete method to find the eccentricity of \(E\). Uses \(b^2 = a^2\left(1 - e^2\right)\) to find a value for \(e\) or \(e^2\)
Ignore references to \(\pm\) for the Ms. Note that identifying \(a\) and \(b\) correctly is part of the method.

dM1: Dependent on the previous method mark. Finds the \(x\) coordinate of the focus of \(E\), 5 multiplied by their value of \(e\) (or implied if working with squared forms).

M1: A complete method to find the value of \(a\). Finds the reciprocal of their eccentricity of \(E\) and uses their \(x\) coordinate of the focus of \(E\) divided by their reciprocal of \(e\). Note they may use \(e^2 = 1 + \dfrac{b^2}{a^2}\) and \(x_F = \sqrt{a^2 + b^2}\) for the hyperbola to form and solve suitable equations.
So for example \(a^2 + b^2 = \text{“}4\text{”}^2\) and \(\text{“}\dfrac{5}{4}\text{”}^2 = 1 + \dfrac{b^2}{a^2}\) leading to a value for \(a\) or \(b\).

A1: Correct value of \(a\) or \(b\). Accept as a decimal.

M1: A complete method to find a value of \(b\) or \(a\) if \(b\) found first. Uses \(b^2 = a^2\left(e^2 - 1\right)\) with their values for \(e\) and \(a\) to find a value for \(b\) or vice versa, allowing slips in substitution as long as the correct formula is clear.

A1: Correct values for both \(a\) and \(b\). Accept as a decimals.

Notes

M1: Uses the correct eccentricity formula and the given equation to find a value for \(e\) or \(e^2\).

A1: Correct exact value \(\dfrac{\sqrt{7}}{4}\), must be simplified. Must reject the negative value, so A0 if \(-\dfrac{\sqrt{7}}{4}\) also included.

Notes

M1: Starts the process of establishing the gradient of the normal by adopting a suitable method to differentiate such as using implicit differentiation or parametric differentiation to obtain an expression for \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) in terms of \(\theta\). The substitution for \(x\) and \(y\) may happen later, but must be in terms of \(\theta\) when used in the equation for normal.

A1: Correct unsimplified \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) in terms of \(\theta\). May be implied (if substitution happens later).

M1: Applies the negative reciprocal to their gradient and uses this and the coordinates of \(P\) to establish an equation of the normal. If using \(y = mx + c\) they must proceed as far as finding \(c\).

A1*: Fully correct proof with sufficient working shown. Must be an intermediate step between the initial equation and the final answer. Do not penalise minor notational slips such as missing some \(\theta\)’s if the working is clear, but award A0 if they are consistently missed.

Notes

B1: Deduces the correct equation of \(OQ\). Need not be simplified - may use point \(Q\).

M1: Proceeds to solve their \(OQ\) line simultaneously with the given \(l_1\) to find either the \(x\) or the \(y\) coordinate of the intersection.

A1: One correct coordinate simplified or unsimplified.

A1: Both correct simplified coordinates. (Accept decimal equivalents.)

Notes

M1: Uses their coordinates of \(R\) (of an appropriate form) correctly to determine the values of “\(a\)” and “\(b\)” for the second ellipse and uses their values with the correct eccentricity formula in an attempt to show that the eccentricities are the same.

A1cso: Must have had correct coordinates in (c), a statement or deduction (or work shown) to verify the locus of \(R\) is an ellipse (conclusion not needed, e.g. accept if the equation is given to show it is an ellipse), and proceeds to show that the value of \(e\) (or \(e^2\) - do not penalise inclusion of the negative value a second time) is the same as that obtained in part (a) or equivalent work. Note that the correct eccentricity can arise from coordinates of \(R\) where the denominator was incorrect, but these will score A0.

Notes

M1: Differentiates implicitly, directly or parametrically to find the gradient at the point \(P\) in terms of \(t\). Allow slips in coefficients, as long as method is clear.

M1: Finds the equation of the tangent at the point \(P\) using their gradient (not reciprocal etc). If using \(y = mx + c\) must proceed to find \(c\) and substitute back in to equation.

A1*: The correct equation for the tangent at the point \(P\) from correct working.

Notes

M1: Finds the new gradient (any method as above) and proceeds to find the equation of the tangent at the point \(Q\). Alternatively replaces \(t\) by \(2t\) in the answer to (a).

A1: Correct equation - any form, need not be simplified and isw after a correct equation.

Notes

M1: Solves their simultaneous equations to find both the \(x\) and \(y\) coordinate for the point \(R\).

A1: Correct point of intersection, it does not need to be simplified.

dM1: Dependent on the first method mark. Multiplies \(x\) by \(y\) to reach a constant.

A1: Shows that \(xy = 32\) and hence rectangular hyperbola

Notes

M1: Uses “\(b^2 = a^2\left(1 - e_1^2\right)\)” with values for \(a\) and \(b\) to find a value for \(e_1\) or \(e_1^2\). They may just call it \(e\) and will likely use \(a\) and \(b\) before substituted, which is fine. The formula must be correct but allow slips with \(a\) and \(b\).

A1: Correct exact value for \(e_1\) or \(e_1^2\). Note: allow M1A1 here if the relevant work is seen in (b).

dM1: Dependent on previous method mark. Uses \(e_1 \times e_2 = 1\) with their \(e_1\) or \(e_1^2\) to find an expression between \(a\) and \(b\). May find an expression for \({e_2}^{(2)}\) and apply \(e_1 \times e_2 = 1\) directly or may first substitute as per scheme. Any full method.

SC: Allow M0A0dM1A0 if \(b^2 = a^2\left(1 - e_1\right)\) and \(b^2 = a^2\left(e_2 - 1\right)\) are used in an otherwise correct process.

A1*: Achieves \(a^2 = 3b^2\) with at least one intermediate unsimplified equation in \(a\) and \(b\) cso

Notes

M1: Uses/implies \(x\) coordinate of focus for the ellipse is \(4 \times\) their \(e_1\)

M1: For a full process to find values for \(a\) and \(b\) or their squares. E.g. for focus of hyperbola \(x = a \times\) their \(e_2 = \dfrac{a}{e_1}\) sets equal to \(4e_1\) and solves for \(a\) then attempting to use \(a^2 = 3b^2\) to obtain \(b^2\) (or \(b\)). Other methods are possible.

A1: Deduces the correct equation for the hyperbola.

Notes

B1: Correct equations for the asymptotes of \(H\) seen or implied, any form and need not be simplified.

Notes

Note M1 M1 A1 A1 on ePen

M1: A correct method to find the gradient of the parabola.

M1: Finds the gradient of the normal and sets their normal gradient equal to their asymptote gradient to obtain at least one point on \(C\) where normal is parallel to an asymptote

M1: Finds the equation of either \(l_1\) or \(l_2\)

A1: Correct equation for each normal, \(y + 8 = \dfrac{1}{2}(x - 2)\) and \(y - 8 = -\dfrac{1}{2}(x - 2)\). Ignore labelling.

Notes

M1: Substitutes for \(x\) or \(y\) into the equation of the hyperbola and solves for their variable.

A1: Achieves one correct coordinate \(x = 10\) and \(y = \pm 4\)

dM1: Dependent on previous method. Correct method for the area of their triangle e.g, \(\dfrac{1}{2} \times\) their \(10 \times\) twice their 4 or equivalent determinant methods.

A1: Area is 40. Correct answer only.

Notes

M1: Uses \(b^2 = a^2\left(1 - e^2\right)\) to obtain a value of \(e\) (allow if \(-\dfrac{2}{3}\) also given)

dM1: Uses \(a = 6\) and their value of \(e\) with \(0 \lt e \lt 1\), to find at least one focus using \(\left((\pm)ae, 0\right)\)

A1: Correct foci – both required, including \(y\) coordinates.

Alternative

M1: Sets up an equation using total distance from foci to point on ellipse \(= 2a\)

dM1: Solves to find a value for the \(x\) coordinate of the foci

A1: Correct foci – both required, including \(y\) coordinates.

Notes

M1: Uses \(x = (\pm)\dfrac{a}{e}\) with \(a = 6\) and their \(e\) to attempt directrices.

A1: Correct directrices, both required and no other lines

Notes

B1: Deduces gradient is \(\dfrac{2}{p + q}\). May be implied by correct simplification of equation if the unsimplified form is used to start with.

M1: Correct method for the equation of the line (gradient need not be simplified/correct for this method, as long as it is clearly an attempt at the gradient).

A1*: Completes to the correct equation with no errors seen.

Notes

M1: A correct overall method – must find both normals and attempt to solve simultaneously. They do not need to reach \(x =\) or \(y =\) as long as they have eliminated one variable.

B1: Correct gradient of normal found from any correct method or just stated.

M1: A full correct method to find the equation of at least one of the normals with justification of the gradient shown.

A1: Deduces equation of the second normal – so both correct.

M1: Solves the two normal equations simultaneously leading to either \(x = \ldots\) or \(y = \ldots\)

A1*: Need to show that \(\left(9p^3 - 9q^3 + 18p - 18q\right) = 9\left(p^2 + q^2 + pq + 2\right)(p - q)\) leading to correct \(x\) coordinate with no errors seen. This could by long division or factorising.

A1*: Correct \(y\) coordinate with no errors seen.

(Corrected from the printed mark scheme: the factorisation is printed as \(\left(9p^2 + q^2 + pq + 2\right)(p - q)\), in the scheme and in the note; the factor 9 belongs outside the bracket.)

Notes

B1: Uses the condition on \(l\) to establish the relationship between \(p\) and \(q\)

M1: Uses their relationship between \(p\) and \(q\) to simplify the expressions

M1: Any complete method for relating \(x\) and \(y\) independently of \(p\) and \(q\)

A1: \(y^2 = 16(x - 30)\) or \(y^2 = 16x - 480\)

Notes

M1: Uses \(b^2 = a^2(1 - e^2)\) with \(a = 6\) and \(b = 4\) to find a value for \(e^2\) or \(e\).

A1: Correct value for \(e\) or \(e^2\)

A1: Correct foci

Notes

M1: Forms a complete strategy to find the perimeter using general \(P\) and applies the focus directrix property to the sides \(PS\) and \(PS^{\prime}\)

M1: Deduces the length of the two sides adjacent to \(P\) is a constant

B1ft: Uses \(SS^{\prime}\) is twice their \(ae\) from (a)

A1*: Finds the value and makes conclusion that perimeter is constant for any \(P\) on \(E\)

Alternative

Scheme	Marks	AO
Perimeter \(= PS + PS^{\prime} + SS^{\prime}\) \(PS = \sqrt{\left(2\sqrt{5} - 6\cos\theta\right)^2 + (4\sin\theta)^2} = \ldots\left\{6 - 2\sqrt{5}\cos\theta\right\}\) \(PS^{\prime} = \sqrt{\left(2\sqrt{5} + 6\cos\theta\right)^2 + (4\sin\theta)^2} = \ldots\left\{6 + 2\sqrt{5}\cos\theta\right\}\)	M1	3.1a
\(PS + PS^{\prime} = \left(6 - 2\sqrt{5}\cos\theta\right) + \left(6 + 2\sqrt{5}\cos\theta\right) = 12\)	M1	2.2a
\(\ldots + 2 \times 2\sqrt{5}\)	B1ft	1.1b
\(= 12 + 4\sqrt{5}\) hence perimeter is constant for any \(P\) on \(E\). *	A1*	2.1
	(4)

M1: Forms a complete strategy to find the perimeter using general \(P\). Finds the lengths of \(PS\) and \(PS^{\prime}\) using Pythagoras theorem and the general coordinate \((6\cos\theta, 4\sin\theta)\)

M1: Deduces the length of the two sides adjacent to \(P\) is a constant, using trig identities.

B1ft: Uses \(SS^{\prime}\) is twice their \(ae\) from (a)

A1*: Finds the value and makes conclusion that perimeter is constant for any \(P\) on \(E\)

(Corrected from the printed mark scheme: the second line of the alternative ends “\(= B\)”; the sum is \(12\).)

Note: Using the property that \(PS + PS^{\prime} = 2a\) both method marks may be awarded as long as a reason is given e.g. definition/property of an ellipse

Notes

M1: Attempts to solve the problem by using differentiation to obtain an expression for \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) in terms of \(\theta\). Allow this mark for \(\dfrac{x^2}{16} - \dfrac{y^2}{9} = 1 \Rightarrow \alpha x - \beta yy^{\prime} = 0 \Rightarrow y^{\prime} = \ldots\) or an attempt to differentiate \(x\) and \(y\) wrt \(\theta\) and then \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{\mathrm{d}y}{\mathrm{d}\theta} \div \dfrac{\mathrm{d}x}{\mathrm{d}\theta} = \ldots\)

M1: Correct straight line method using the coordinates of \(P\) and their gradient in terms of \(\theta\)
Allow the results for the first 2 M marks to be “quoted”, but any statements must be correct to score the marks.

A1: Uses \(y = 0\) to deduce the correct coordinates (or value of \(x\)) for the point \(A\). Allow in any form, simplified or unsimplified (e.g. unsimplified: \(x = 4\cosh\theta - 4\tanh\theta\sinh\theta\))

B1: Deduces that the equation of \(l_2\) is \(x = 4\) (may be implied by \(x = 4\) used to find \(y\) coordinate of \(B\))

M1: Realises that \(x = 4\) is all that is needed for the second line and substitutes this into the first line in order to find the point \(B\)

A1: Deduces the correct coordinates or \(y\) value for \(B\)
(e.g. unsimplified \(y = \dfrac{3}{\tanh\theta} - \dfrac{3\cosh\theta}{\tanh\theta} + 3\sinh\theta\))

M1: Uses a correct method for the midpoint of \(AB\) (coordinates must be the right way round). This may be seen as the coordinates written separately e.g. \(x = \ldots\), \(y = \ldots\)

M1: Having found the midpoint, identifies a correct strategy that will enable a Cartesian equation to be found. E.g. find \(\cosh\theta\) in terms of \(x\) and substitutes into \(y\) or \(y^2\) to obtain an equation in terms of \(y\) and \(x\) only. Mark positively here, so allow the mark if the candidate makes progress in eliminating \(\theta\) even if there are slips in the working.

A1*: Obtains the printed answer with no errors

Alternative for the previous 2 marks: Substitutes the coordinates of their midpoint into both sides of the given equation in an attempt to show they are equal. Again mark positively but having made the substitution, some progress needs to be made in showing that both sides are equal. For this method there must be a minimal conclusion for the A1 e.g. tick, hence true etc.

Note that these 2 marks can also be attempted by expressing the midpoint in terms of exponentials – if you are in doubt whether to award marks seek advice from your Team Leader.

M1: For \(p = 2\) or \(q = 4\)

A1: For \(p = 2\) and \(q = 4\)

Notes

M1: A complete method for finding the \(x\) coordinate of the focus using a correct eccentricity formula to find a value for e and then calculating \(4e\)

M1: Completes the problem by subtracting 4 from the \(x\) coordinate of the focus

A1*: Correct answer

If you come across correct attempts using Pythagoras to prove the result send to review.

Notes

M1: Attempts to solve the problem by using differentiation to obtain an expression for \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) in terms of \(p\) or \(q\).
See scheme for requirements for this mark depending on the method chosen.
(Can be implied by a correct expression)

M1: Correct straight line method to find the equation of the tangent using \(P\) or \(Q\).
If using \(y = mx + c\), must reach as far as \(c = \ldots\)

A1: Obtains a correct general tangent at \(P\) or \(Q\) or both
Note that if a correct tangent equation is quoted, the first 3 marks are available

M1: Uses \(x = -28\) and \(y = 6\) with the values correctly placed in one of their tangent equations and attempts to solve the resulting 3TQ to obtain 2 values for \(p\) (or \(q\)).
An alternative approach for this mark is to obtain equations for both tangents and solve simultaneously to obtain the coordinates for the intersection and then to use \(x = -28\) and \(y = 6\) to find values for \(p\) and \(q\). Note that a calculator may be used for the simultaneous equations but answers must be correct for their equations if no working is shown.

A1: Correct values

A1: Deduces the correct coordinates of \(P\) and \(Q\)

M1: Completes the problem by using a suitable complete correct method for finding the area of \(PQR\) – See examples – there will be others – in general, score M1 for a correct triangle area method for their values

A1*: Correct area. Allow this mark even if the candidate reverts to decimals within their solution, providing all the working is correct.

Generally, using midpoints of sides is unlikely to be successful, however, the line from \(R\) to the midpoint of \(PQ\) is horizontal so this is a correct approach:

Midpoint: \(\left(\dfrac{49 + 16}{2}, \dfrac{28 - 16}{2}\right) = \left(\dfrac{65}{2}, 6\right) \Rightarrow \text{Area} = \dfrac{1}{2}\left(\dfrac{65}{2} + 28\right) \times 44 = 1331\)

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

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.

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

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

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).

Notes

M1: Solves \(\dfrac{\mathrm{d}T}{\mathrm{d}m} = 0\) to obtain a value for \(m\)

A1: cao

Alternative for (e)

M1: Writes \(T\) as \(\dfrac{(5m - 4)^2 + \ldots}{2m}\) and realises minimum when \((5m - 4) = 0\)

A1: cao

Notes

M1: Uses \(S(a, 0)\) and \(D\left(\text{their "}-a\text{"}, \dfrac{24a}{5}\right)\) to find an expression for the gradient of \(l\) or applies the formula \(\dfrac{y - y_1}{y_2 - y_1} = \dfrac{x - x_1}{x_2 - x_1}\). Can be un-simplified or simplified.

A1 *: Correct solution only leading to \(12x + 5y = 12a\). No errors seen.

ALT (a)

M1: Uses \(S(a, 0)\) and \(D\left(\text{their "}-a\text{"}, \dfrac{24a}{5}\right)\) to find 2 simultaneous equations and solves to achieve \(c = \ldots, m = \ldots\)

A1*: Correct solution only leading to \(12x + 5y = 12a\)

Notes

M1: Attempts to find the gradient of \(SP\)

M1: Some evidence of applying \(m_1m_2 = -1\)

A1: Correct 3TQ in terms of \(k\) in any form.

M1: Attempt to solve their 3TQ for \(k\)

M1: Uses their \(k\) to find \(P\)

A1: \((25a, 10a)\)

ALT 1 (b)

M1: \(y - 0 = m_{SP}(x - a)\)

M1: \(m_{SP} = -\dfrac{1}{(-\frac{12}{5})}\ \left\{= \dfrac{5}{12}\right\}\)

Can sub for \(x\) and achieve \(\dfrac{12}{5}y + a\)

A1: Correct 3TQ in terms of \(a\) and \(x\) or \(5y^2 - 48ay - 20a^2 = 0\)

M1: Attempt to solve their 3TQ for \(x\)

M1: Uses their \(x\) to find \(y\)

A1: \((25a, 10a)\)

ALT 2 (b)

M1: Subs \(S\) into \(y = m_{SP}x + c\) to find \(c\)

M1: Some evidence of applying \(m_1m_2 = -1\)

A1: Correct 3TQ in terms of \(k\)

Notes

M1: \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \pm kx^{-2}\) or \(y + x\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0\) or \(\dfrac{\text{their } \frac{\mathrm{d}y}{\mathrm{d}t}}{\text{their } \frac{\mathrm{d}x}{\mathrm{d}t}}\)

A1: \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = -\dfrac{1}{t^2}\)

M1: Applies \(y - \dfrac{c}{t} = \left(\text{their } m_T\right)(x - ct)\) where their \(m_T\) has come from calculus

At least one line of working.

A1 cso *: Correct solution.

Notes

M1: Substitutes \(\left(-\dfrac{8c}{5}, \dfrac{3c}{5}\right)\) into tangent.

A1: Correct 3TQ in terms of \(t\). Can include uncancelled \(c\).

M1: Attempt to solve their 3TQ for \(t\)

M1: Uses one of their values of \(t\) to find \(A\) or \(B\)

A1: Correct coordinates. Condone \(A\) and \(B\) swapped or missing.

ALT 1 (b)

M1: Substitutes \(\left(ct, \dfrac{c}{t}\right)\) into their \(y - \dfrac{3c}{5} = -\dfrac{1}{t^2}\left(x - -\dfrac{8c}{5}\right)\)

A1: Correct 3TQ in terms of \(t\). Can include uncancelled \(c\).

ALT 2 (b)

M1: Substitutes \(A\) and \(B\) into the equation of the tangent, solves for \(x\) and \(y\)

A1: Correct 3TQ in terms of \(t_1\) or \(t_2\). Can include uncancelled \(c\).

Notes

M1: \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \pm k\,x^{-\frac{1}{2}}\) or \(ky\dfrac{\mathrm{d}y}{\mathrm{d}x} = c\) or \(\dfrac{\text{their } \frac{\mathrm{d}y}{\mathrm{d}q}}{\text{their } \frac{\mathrm{d}x}{\mathrm{d}q}}\)

A1: \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{q}\)

dM1: Applies \(y - 2aq = \left(\text{their } m_T\right)\left(x - aq^2\right)\) or \(y = \left(\text{their } m_T\right)x + c\) and an attempt to find \(c\) with gradient from calculus.

A1 *: cso

Notes

M1: Substitutes \(x = -\tfrac{1}{4}a\) and \(y = 0\) into T

A1: \(q = \dfrac{1}{2}\) oe

M1: Substitutes their "\(q = \tfrac{1}{2}\)" and \(x = -a\) in T or finds \(y_D = \dfrac{1}{q}\left(-a + aq^2\right)\)

A1: \(D\left(-a, -\tfrac{3}{2}a\right)\) o.e.

Notes

M1: Applies \(\dfrac{1}{2}\left(\text{their } |FX|\right)\left(\text{their } |y_D|\right)\). If their \(\left|y_D = \dfrac{1}{q}\left(-a + aq^2\right)\right|\) then require an attempt to sub for \(q\) to award M.

A1 cso: \(\dfrac{15a^2}{16}\) or \(0.9375a^2\)

Do not award M1 if area of wrong triangle found e.g. \(\dfrac{1}{2}.2a.\dfrac{3a}{2} = \dfrac{3a^2}{2}\)

(c) Way 2

M1: A correct attempt to apply the shoelace method.

A1cao: \(\dfrac{15a^2}{16}\) or \(0.9375a^2\)

(c) Way 3

(c) Way 4

M1: Uses Area \(= \dfrac{1}{2}ab\sin C\)

Notes

B1: Coordinates for either \(P\) or \(Q\) are correctly stated. (Can be implied).

M1: Finds the gradient of the chord \(PQ\) with \(\dfrac{y_2 - y_1}{x_2 - x_1}\) then uses in \(y = -\dfrac{1}{m}x\). Condone incorrect sign of gradient.

A1 oe: \(y = \dfrac{1}{2}x\) or \(2y = x\)

Notes

B1 oe: Correct Cartesian equation. Accept \(\dfrac{4}{y} = \dfrac{x}{4}\) or \(xy = 4^2\)

Notes

M1: Attempts to substitute their \(l\) into either their Cartesian equation or parametric equations of \(H\)

A1: At least one set of coordinates (simplified or un-simplified) or \(x = \pm 4\sqrt{2},\ y = \pm 2\sqrt{2}\)

A1: Both sets of simplified coordinates. Accept written in pairs as \(x = 4\sqrt{2},\ y = 2\sqrt{2}\); \(x = -4\sqrt{2},\ y = -2\sqrt{2}\)

Notes

M1: Correct attempt at \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) using implicit or parametric or explicit differentiation
\(\left(ax + by\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0 \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = \ldots,\ \dfrac{\mathrm{d}y}{\mathrm{d}x} = \pm\dfrac{a\cos\theta}{b\sin\theta},\ \dfrac{\mathrm{d}y}{\mathrm{d}x} = ax\left(1 - bx^2\right)^{-\frac{1}{2}}(oe) \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = \ldots\right)\)

A1: Correct tangent gradient in terms of \(\theta\). May be implied in their attempt the normal gradient.

M1: Correct perpendicular gradient rule. May be awarded if working in terms of \(x\) and \(y\).

M1: Correct straight line method for the normal using a “changed” \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) in terms of \(\theta\) which must have come from calculus. If using \(y = mx + c\), must reach as far as \(c = \ldots\)

A1*: Correct completion to printed answer with no errors.

Note that if the candidate uses e.g \(y - 5\sin\theta = -\dfrac{36y}{25x}(x - 6\cos\theta)\) before introducing \(\theta\), the final mark can be withheld.

Notes

M1: Uses the correct eccentricity formula to obtain a value for \(e\) or \(e^2\). Ignore \(\pm\) values for e.

B1: Correct \(x\) coordinate for \(Q\)

M1: Attempts \(\dfrac{\text{their } OQ}{\text{their } OR}\). May be implied by their ratio.

A1: Correct completion with no errors to obtain \(\dfrac{11}{36}\) both times.

Ignore any references to the foci or directrices but the final mark can be withheld if there are any incorrect statements such as e.g. using \(\cos\theta = 1\) in their ratio.

Notes

B1: Any correct expression for gradient of tangent
M1: Substitutes values into derived expression using calculus to give gradient of normal at \(P\)
A1: cao. Can be implied by use in equation of a straight line
M1: Use of formula for the equation of a straight line with their changed gradient
A1: cso

Notes

M1: Replaces \(x\) by \(-y\) or \(y\) by \(-x\)
M1: Factorises \((1 - p^4)\) to simplify answer in first variable
A1 cso: Obtains both \(x\) and \(y\)
ALT (b) Accept Verification.
M1: Substitutes the coordinates of \(A\) into the equation of the normal
M1: Substitutes the coordinates of \(A\) into both the normal and \(y = -x\).
A1 cso: No errors seen

Notes

B1: Correct \(x\)- coordinate of midpoint (may be implied) and correct \(y\) coordinate, accept equivalent forms
M1: Puts their \(y = 0\) and finds value for \(p\) to use in \(x =\)
A1: \(+\dfrac{15}{4}\sqrt{2}\) or equivalent only

Notes

M1: Correct gradient method. Finds \(\dfrac{\mathrm{d}y}{\mathrm{d}\theta} = p\sec^2\theta\) and \(\dfrac{\mathrm{d}x}{\mathrm{d}\theta} = q\sec\theta\tan\theta\) and uses \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{\mathrm{d}y}{\mathrm{d}\theta} \times \dfrac{\mathrm{d}\theta}{\mathrm{d}x}\) or differentiates implicitly to give \(px + qy\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0\) and substitutes for \(y\) and \(x\) to find \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) or differentiates explicitly to give \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = px\left(qx^2 - 1\right)^{-\frac{1}{2}}\) and substitutes for \(x\)

A1: Correct derivative in terms of trig. functions, e.g. \(\dfrac{3\sec^2\theta}{4\sec\theta\tan\theta},\ \dfrac{8\sec\theta}{16} \times \dfrac{9}{6\tan\theta}\) Does not need to be simplified.

M1: Correct perpendicular gradient rule. Does not need to be simplified.

M1: Correct straight line method using a gradient (does not need to be simplified) in terms of \(\theta\) that has come from calculus and is not the tangent gradient. If they use \(y = mx + c\) then they must reach as far as finding \(c\).

A1*: Correct proof with no errors and one intermediate step from the previous line. Allow \(25\tan\theta = 3y + 4x\sin\theta\)

Notes

M1: Use of the correct eccentricity formula to obtain a value for \(e\)

A1: Correct value for \(e\). Ignore \(\pm\)

A1: Correct value for \(\dfrac{a}{e}\) Ignore \(\pm\)

M1: Substitutes \(\theta = \dfrac{\pi}{4}\) into the given normal equation and uses their positive directrix equation to obtain an equation in \(y\) or in \(y\) and e only.

B1: \(a = \dfrac{25}{3}\) oe or \(b = -\dfrac{32}{15}\) oe

B1: \(a = \dfrac{25}{3}\) oe and \(b = -\dfrac{32}{15}\) oe

Special Case: If the correct form of the answer is never seen but it appears correctly as a single fraction, allow B1B0 e.g. \(y = \dfrac{125 - 32\sqrt{2}}{15}\)

Notes

B1: Correct statement for gradient (isw) B1: \(\dfrac{2}{p + q}\) - can be seen later in the solution.
M1: use of a correct formula for a line equation through \(P\) or \(Q\) with their gradient. Must be finding a chord, not a tangent or a normal.
A1: for a correct line equation with simplified gradient in any equivalent form
A1: cso (as given answer)

Notes

B1: For using \((a, 0)\) to show that \(pq = -1\)

Notes

M1: Use calculus to find an expression for \(\mathrm{d}y/\mathrm{d}x\) and substitute coordinates of \(P\). They may use chord gradient and let \(p\) tend to \(q\).

Notes

B1: \(1/q\) seen B1: \(\dfrac{1}{p} \times \dfrac{1}{q} = -1\) or \(\dfrac{1}{p} = -\dfrac{1}{\frac{1}{q}}\) or \(\dfrac{1}{q} = -\dfrac{1}{\frac{1}{p}}\) and at least words in bold with no errors seen.

Notes

M1: Uses a correct eccentricity formula to find a value for e or e²

A1: \(e^2 = \frac{3}{4}\) or \(e = \frac{\sqrt{3}}{2}\) (allow \(e = \pm\frac{\sqrt{3}}{2}\))

B1: Both correct as coordinates

B1: Both directrices correct seen as equations. Accept un-simplified e.g. \(x = \pm\dfrac{2}{\sqrt{3}/2}\)

Notes

M1: Use of definition of ellipse for either \(PF_1\) or \(PF_2\)

dM1: \((\text{their } e) \times 2\left(\text{their } \frac{4}{\sqrt{3}}\right)\) Dependent on the previous method mark

A1: \(\frac{\sqrt{3}}{2} \times \left(2 \times \frac{4}{\sqrt{3}}\right)\)

A1**: cso

(b) Alternative 1: Using \(P(2\cos\theta, \sin\theta)\) (Must be of this form)

M1: Correct use of Pythagoras for either \(PF_1\) or \(PF_2\)

dM1: Obtains both \({PF_1}^2 = \left(\sqrt{3}\cos\theta - \sqrt{p^2 + 1}\right)^2\) and \({PF_2}^2 = \left(\sqrt{3}\cos\theta + \sqrt{p^2 + 1}\right)^2\) where \(p\) is the \(x\)-coordinate of a focus. Dependent on the previous method mark

A1: \(2 - \sqrt{3}\cos\theta + 2 + \sqrt{3}\cos\theta\). Note that if \(\sqrt{3}\cos\theta - 2\) is obtained correctly, it must become \(2 - \sqrt{3}\cos\theta\) to score any A marks

A1**: cso

(b) Alternative 2: Using \(P\left(x, \sqrt{\dfrac{4 - x^2}{4}}\right)\) (Must be of this form) or \(P\left(\sqrt{4 - 4y^2}, y\right)\)

M1: Correct use of Pythagoras for either \(PF_1\) or \(PF_2\)

dM1: Obtains both \({PF_1}^2 = \left(\dfrac{\sqrt{3}}{2}x - \sqrt{p^2 + 1}\right)^2\) and \({PF_2}^2 = \left(\dfrac{\sqrt{3}}{2}x + \sqrt{p^2 + 1}\right)^2\) where \(p\) is the \(x\)-coordinate of the foci. Dependent on the previous method mark

A1: \(2 - \dfrac{\sqrt{3}}{2}x + 2 + \dfrac{\sqrt{3}}{2}x\)

A1**: cso

Notes

M1: Substitutes the equation of a straight line with gradient \(m\) into the equation of the ellipse

A1: Correct quadratic in \(x\) with terms collected

M1: Attempts \(\dfrac{1}{2}\)(sum of roots)

A1: Correct expression for \(c\) in terms of \(m\) and \(x\)

ddM1: Substitutes back into \(y = mx + c\) Depends on both previous method marks

A1: Correct equation

Or for last 3 marks

A1: Correct \(y\)-coordinate in terms of \(m\) and \(c\).

ddM1: Obtains \(y\) in terms of \(x\) and \(m\) Depends on both previous method marks

A1: Correct equation

Alternative: Using factor formulae

Let ends of the chord be \((2\cos\alpha, \sin\alpha)\) and \((2\cos\beta, \sin\beta)\) (Must be of this form)

Special Case

Notes

M1: Uses distance between two points or states perpendicular distance from \(P\) to directrix required.
B1: States or uses focus at \((3,0)\) or focus at \(a = 3\) or directrix as \(x = -3\)
A1: cso

Notes

M1: Calculus method for finding gradient and substitutes \(x\) value at either point
A1: Either correct. Accept unsimplified.
A1: One equation of tangent correct. B1: Both correct
M1: Eliminate \(x\) or \(y\). A1: Obtain first variable
M1: Substitute or eliminate again. A1: Both variables correct in simplest form as above.

Notes

M1: Find their \(SR^2 = (3 - 3pq)^2 + (3p + 3q)^2\ (= 9 + 9p^2q^2 + 9p^2 + 9q^2)\)
M1: Find their \(SP.SQ = 3(1 + p^2)\ 3(1 + q^2)\ (= 9 + 9p^2q^2 + 9p^2 + 9q^2)\)
A1: Deduce equal after no errors seen. Concluding statement required cso.

Notes

B1: Both required. Accept \(y = \pm x\) and \(x = \pm y\)

Notes

B1: Correct gradient

M1: Correct straight line method. For \(y = mx + c\) method, \(c\) must be found

A1* cso: Obtains the printed answer with at least one intermediate step.

Notes

B1: All four values correct. May be in exponential form e.g. \(\left(e^t,\ e^t\right)\) and \(\left(e^{-t},\ -e^{-t}\right)\)

M1: Correct attempt at \(X\) or \(Y\). May be in exponential form e.g. \(\left(\dfrac{e^t + e^{-t}}{2},\ \dfrac{e^t - e^{-t}}{2}\right)\)

A1cso: Obtains \(X = \cosh t\) and \(Y = \sinh t\). May be shown using exponentials as above.

Notes

M1: Correct triangle area method

A1: Obtains an area of 1

A1ft: Concludes independence of \(t\) having obtained a constant area. Conclusion must include the word independent (or not dependent) (but not e.g. just QED)

Alternative area method

If \(A\left(\frac{1}{\cosh t}, 0\right)\) is the intersection of \(QR\) with the \(x\)-axis

Area OAR + Area OAQ \(= \dfrac{1}{2} \times \dfrac{1}{\cosh t} \times \dfrac{1}{\cosh t - \sinh t} + \dfrac{1}{2} \times \dfrac{1}{\cosh t} \times \dfrac{1}{\cosh t + \sinh t}\)

\(= \dfrac{1}{2} \times \dfrac{1}{\cosh t} \times \left(\dfrac{1}{\cosh t - \sinh t} + \dfrac{1}{\cosh t + \sinh t}\right) = \dfrac{1}{2\cosh t} \times 2\cosh t = 1\)

Notes

M1: Differentiates to obtain \(\dfrac{k}{x^2}\) and substitutes \(x = 6\)
or uses implicit differentiation \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = -\dfrac{y}{x}\) and substitutes \(x\) and \(y\)
or uses parametric differentiation \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = -\dfrac{1}{t^2}\) and substitutes \(t = 2\)
A1: For grad of tangent – accept any equivalent i.e. \(-0.25\) etc
M1: Uses negative reciprocal of their gradient.
dM1: \(y - y_1 = m(x - x_1)\) with \(\left(6, \dfrac{3}{2}\right)\) or \(y = mx + c\) and sub \(\left(6, \dfrac{3}{2}\right)\) to find \(c=\).
A1: cso: Correct answer with no errors seen in the solution.

Notes

M1: Obtains equation in one variable, \(x\) or \(y\)
A1: Correct value of \(x\) or correct value of \(y\)
M1: Finds second coordinate using \(xy = 9\) or solving second quadratic or equation of the normal
A1: Correct coordinates that can be written as \(x = \ldots, y = \ldots\)

Notes

(d) Way 2

Notes

This is cso – there must no errors in applying the factor formulae and sufficient working must be shown to justify the printed answer but allow \(\cos\beta\cos\frac{1}{2}(\alpha + \beta) + \sin\beta\sin\frac{1}{2}(\alpha + \beta) = \cos\dfrac{\alpha - \beta}{2}\)

Notes

Notes

Alt Method: An attempt to eliminate either \(x\) or \(y\) from \(xy = 9\) and \(6y = 4x - 15\)

1^st M1: A full method to obtain a quadratic equation in either \(x\) or \(y\).
1^st A1: For either \(4x^2 - 15x - 54 = 0\) or \(6y^2 + 15y - 36 = 0\) or equivalent.
2^nd M1: A valid attempt at solving the quadratic.
2^nd A1: For either \(x = 6, -\dfrac{9}{4}\) or \(y = \dfrac{3}{2}, -4\)
3^rd A1: Both \(\left(6, \dfrac{3}{2}\right)\) and \(\left(-\dfrac{9}{4}, -4\right)\).

Notes

Special Case: In (b) they assume the locus is a circle and find the intercepts on the \(x\)-axis as (1, 0) and (7, 0) and hence deduce the centre (4, 0) and radius 3. This approach scores no marks in (b) but allow recovery in (c).

Notes

Alternative

Notes

M1: \((x = )(\pm)\dfrac{a}{e}\)   \(\pm\) not needed for this mark nor is \(x\) and even allow \(y = (\pm)\dfrac{a}{e}\) here – just look for use of \(\dfrac{a}{e}\) with numerical \(a\) and \(e\).

A1: \(x = \pm\dfrac{144}{13}\) oe but must be an equation (Do not allow \(x = \pm\dfrac{12}{13/12}\))

If they use the eccentricity equation for the ellipse \((b^2 = a^2(1 - e^2))\) allow the M’s

Notes

Alternatives:
First M1 for \(ky\dfrac{\mathrm{d}y}{\mathrm{d}x} = 36\) or for
\(x = 9t^2,\ y = 18t \to \dfrac{\mathrm{d}x}{\mathrm{d}t} = 18t,\ \dfrac{\mathrm{d}y}{\mathrm{d}t} = 18 \to \dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{t}\)

(a) First M for \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = ax^{-\frac{1}{2}}\),

Second M for substituting \(x = 4\) (or \(y = 12\) or \(t = 2/3\) if alternative used) into their gradient and applying negative reciprocal.

First A for \(-\dfrac{2}{3}\)

Third M for \(y - y_1 = m(x - x_1)\) or \(y = mx + c\) and attempt to substitute a changed gradient AND (4,12)

Second A for \(3y + 2x = 44\) or any equivalent equation

Notes

(b) M for Area\(= \dfrac{1}{2}\) base x height and attempt to substitute including their numerical ‘(22-9)’ or equivalent complete method to find area of triangle \(PSN\).

Notes

Alternatives for first M1 A1 in part (a)

(a) First M for attempt at explicit, implicit or parametric differentiation not using \(p\) or \(q\) as an initial parameter, first A for \(\dfrac{-1}{p^2}\) or equivalent. Quoting gradient award first M0A0. Second M for using \(y - y_1 = m(x - x_1)\) and attempt to substitute or \(y = mx + c\) and attempt to find c; gradient in terms of \(p\) only and using \(\left(5p, \dfrac{5}{p}\right)\), second A for correct solution only.

Notes

(c) First M for eliminating \(x\) and reaching \(y = \mathrm{f}(p, q)\), second M for eliminating \(y\) and reaching \(x = \mathrm{f}(p, q)\), both As for given answers. Minimum amount of working given in the main scheme above for 4/4, but do not award accuracy if any errors are made.

Notes

(d) First M for use of \(\dfrac{y_2 - y_1}{x_2 - x_1}\) and substituting, first A for \(\dfrac{-1}{pq}\) or unsimplified equivalent .

Second M for their product of gradients\(= -1\) (or equating equivalent gradients of \(ON\) or equating equivalent gradients of \(PQ\)), second A for correct answer only.

Notes

(a) Candidates who derive \(x + t^2y = 2ct\), by stating that \(m_T = -\dfrac{1}{t^2}\), with no justification score no marks in (a).

Notes

Do not allow the \(x\) and \(y\) coordinates of P to be used for the dimensions of the triangle.

Notes

a1M1: Finding gradient in terms of \(\theta\). Must use calculus.

a1A1: cao

a2M1: Finding equation of tangent

a2A1: cso (answer given). Need to get \(\cos^2\theta + \sin^2\theta\) on the same side.

Notes

b1M1: Finding gradient and equation of tangent, or setting \(a = b\).

b1A1: cao need not be simplified.

Notes

c1M1: As scheme

c1A1: \(x = \dfrac{a}{\cos\theta}\), need not be simplified.

c1B1: \(y = 0\), need not be simplified.

Notes

d1B1: Identifying locus as \(y = 0\) or real/’\(x\)’ axis.

d2B1: Depends on previous B mark, identifies correct parts of \(y = 0\). Condone use of strict inequalities.

Notes

Allow any equivalent form e.g. \(k(12x - 5y - 24) = 0\) where \(k\) is an integer

Notes

a1M1: Uses \(b^2 = a^2(e^2 - 1)\) to get \(e > 1\)

a1A1: cao

a2M1: Uses \(ae\)

Notes

b1M1: Uses \(\dfrac{a}{e}\)

b1A1: cso for both foci and both directrices. Must have both of the 2 previous M marks may be implicit.

Notes