FP1-4 Coordinate systems (Further Pure 1) — revision question pack

4 specification points · notes, questions, answers and worked methods

Checked against Edexcel 9FM0 section FP1-4. Review basis: the qualification registry sourced from the Pearson Edexcel Level 3 Advanced GCE in Further Mathematics (9FM0) specification; registry verification recorded 17 July 2026.

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FP1-4.1 · Cartesian and parametric equations for the parabola and rectangular hyperbola, ellipse and hyperbola.

Explanation

Four standard conics carry standard parametrisations.
The parabola $y^2=4ax$ is written as $\left(at^2,\,2at\right)$ ; the rectangular hyperbola $xy=c^2$ as $\left(ct,\,\dfrac{c}{t}\right)$ ; the ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ as $\left(a\cos t,\,b\sin t\right)$ ; and the hyperbola $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$ as $\left(a\sec t,\,b\tan t\right)$ or $\left(a\cosh t,\,b\sinh t\right)$ .
To convert a parametrisation to Cartesian form, eliminate the parameter using the identity that matches it: $\cos^2t+\sin^2t=1$ , $1+\tan^2t=\sec^2t$ or $\cosh^2t-\sinh^2t=1$ .
Translating a conic replaces $x$ by $x-p$ and $y$ by $y-q$ throughout, moving the centre or vertex to $(p,q)$ without changing the shape.
Note that $\left(a\cosh t,\,b\sinh t\right)$ traces only the branch with $x\ge a$ , because $\cosh t\ge1$ , whereas the secant-tangent form covers both branches.

Worked example

Find the Cartesian equation of the curve with parametric equations $x=4\sec t$ , $y=3\tan t$ , and state the coordinates of its vertices.

1. $\sec t=\dfrac{x}{4}$ and $\tan t=\dfrac{y}{3}$ .
2.Substituting into $\sec^2t-\tan^2t=1$ gives $\dfrac{x^2}{16}-\dfrac{y^2}{9}=1$ .
3.The vertices are where $y=0$ , that is $x=\pm4$ .

Answer: $\dfrac{x^2}{16}-\dfrac{y^2}{9}=1$ , with vertices $(4,0)$ and $(-4,0)$ .

Common mistakes

Don't fall into the trap of using $\cos^2t+\sin^2t=1$ on a secant-tangent parametrisation, producing an ellipse instead of a hyperbola.
Don't fall into the trap of reading $4a$ as $a$ in $y^2=4ax$ , so every focus and directrix is four times too far out.
Don't fall into the trap of forgetting that $x=a\cosh t$ gives only one branch of a hyperbola.

Exam tip

Write down which Pythagorean-type identity you are about to use before eliminating the parameter; it makes the conic type obvious and stops the classic ellipse-for-hyperbola slip.

Tier 1 · Easy

1.
A parabola has parametric equations $x=5t^2$ , $y=10t$ . Find its Cartesian equation and the value of $a$ in the standard form $y^2=4ax$ .
(3)
(Total for Question 1 is 3 marks)
2.
Find the Cartesian equation of the curve with parametric equations $x=6\cos t$ , $y=4\sin t$ , and state the lengths of its semi-axes.
(3)
(Total for Question 2 is 3 marks)

Tier 2 · Standard

1.
The rectangular hyperbola $xy=16$ is met by the line $y=x-6$ . Find the coordinates of the two points of intersection and the corresponding values of the parameter $t$ in $\left(4t,\,\dfrac4t\right)$ .
(5)
(Total for Question 1 is 5 marks)
2.
Find the Cartesian equation of the curve with parametric equations $x=2+3\cos t$ , $y=-1+5\sin t$ , and state the coordinates of its centre.
(4)
(Total for Question 2 is 4 marks)
3.
A hyperbola has parametric equations $x=3\cosh t$ , $y=4\sinh t$ . Find its Cartesian equation, the equations of its asymptotes, and state which branch the parametrisation traces.
(5)
(Total for Question 3 is 5 marks)

Tier 3 · Hard

1.
The points $P\left(ap^2,2ap\right)$ and $Q\left(aq^2,2aq\right)$ lie on the parabola $y^2=4ax$ . Show that the chord $PQ$ has equation $(p+q)y=2x+2apq$ .
(5)
(Total for Question 1 is 5 marks)
2.
The points $P\left(cp,\dfrac{c}{p}\right)$ and $Q\left(cq,\dfrac{c}{q}\right)$ lie on the rectangular hyperbola $xy=c^2$ . Show that the chord $PQ$ has equation $x+pqy=c(p+q)$ .
(5)
(Total for Question 2 is 5 marks)
3.
Show that the line $y=mx+k$ meets the ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ in two distinct points if and only if $k^2<a^2m^2+b^2$ .
(6)
(Total for Question 3 is 6 marks)
4.
Find the Cartesian equation of the curve with parametric equations $x=3+2\cosh t$ , $y=-1+5\sinh t$ , and state its centre, its asymptotes and the range of values of $x$ that the parametrisation produces.
(6)
(Total for Question 4 is 6 marks)
5.
A hyperbola has asymptotes $y=\pm\dfrac34x$ and passes through the point $\left(8,\,3\sqrt3\right)$ . Find its equation in the form $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$ .
(6)
(Total for Question 5 is 6 marks)

FP1-4.2 · The focus-directrix properties of the parabola, ellipse and hyperbola, including the eccentricity.

Explanation

Every conic can be defined by a focus $S$ , a directrix $\ell$ and an eccentricity $e$ , as the locus of points $P$ with $SP=e\times(\text{distance from }P\text{ to }\ell)$ . For the parabola $y^2=4ax$ , $e=1$ , the focus is $(a,0)$ and the directrix is $x=-a$ .
For the ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ with $a>b$ , $e<1$ and $b^2=a^2\left(1-e^2\right)$ , the foci are $(\pm ae,0)$ and the directrices are $x=\pm\dfrac{a}{e}$ .
For the hyperbola $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$ , $e>1$ and $b^2=a^2\left(e^2-1\right)$ , with the same expressions for the foci and directrices.
The focal distance of a point on an ellipse is $SP=a-ex$ , so $SP+S'P=2a$ ; on a hyperbola it is $SP=ex-a$ on the branch nearer the focus.
Learn which of $1-e^2$ and $e^2-1$ belongs to which curve; that single sign is the most common source of lost marks.

An ellipse with its two foci at x = ±ae and its two directrices at x = ±a/e; SP equals e times the horizontal distance from P to the near directrix.

Worked example

Find the eccentricity, foci and directrices of the ellipse $\dfrac{x^2}{25}+\dfrac{y^2}{16}=1$ .

1.Here $a^2=25$ and $b^2=16$ , so $16=25\left(1-e^2\right)$ .
2.Hence $e^2=\dfrac{9}{25}$ and $e=\dfrac35$ .
3.The foci are $(\pm ae,0)=(\pm3,0)$ and the directrices are $x=\pm\dfrac{a}{e}=\pm\dfrac{25}{3}$ .

Answer: $e=\dfrac35$ , foci $(\pm3,0)$ , directrices $x=\pm\dfrac{25}{3}$ .

Common mistakes

Don't fall into the trap of using $b^2=a^2\left(e^2-1\right)$ for an ellipse, which gives an imaginary eccentricity.
Don't fall into the trap of quoting the directrix as $x=\pm ae$ instead of $x=\pm\dfrac{a}{e}$ .
Don't fall into the trap of assuming $a^2$ is always the denominator under $x^2$ ; when the major axis is vertical the larger denominator sits under $y^2$ and the foci lie on the $y$ -axis.

Exam tip

Write $b^2=a^2\left(1-e^2\right)$ for the ellipse and $b^2=a^2\left(e^2-1\right)$ for the hyperbola at the top of the page; then read off which one the question needs.

Tier 1 · Easy

1.
State the coordinates of the focus and the equation of the directrix of the parabola $y^2=16x$ .
(2)
(Total for Question 1 is 2 marks)
2.
Find the eccentricity and the coordinates of the foci of the ellipse $\dfrac{x^2}{25}+\dfrac{y^2}{16}=1$ .
(3)
(Total for Question 2 is 3 marks)

Tier 2 · Standard

1.
Find the eccentricity, the foci and the directrices of the hyperbola $\dfrac{x^2}{9}-\dfrac{y^2}{16}=1$ .
(4)
(Total for Question 1 is 4 marks)
2.
An ellipse has foci $(\pm4,0)$ and eccentricity $\dfrac23$ . Find its equation in the form $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ .
(4)
(Total for Question 2 is 4 marks)
3.
The point $P$ lies on the parabola $y^2=12x$ and is $7$ units from the focus. Find the coordinates of the two possible positions of $P$ .
(5)
(Total for Question 3 is 5 marks)

Tier 3 · Hard

1.
Using the focus-directrix property, show that for the ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ with foci $S(ae,0)$ and $S'(-ae,0)$ , the sum $SP+S'P$ is $2a$ for every point $P$ on the curve.
(6)
(Total for Question 1 is 6 marks)
2.
Show that the length of the latus rectum of the ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ , the focal chord perpendicular to the major axis, is $\dfrac{2b^2}{a}$ .
(5)
(Total for Question 2 is 5 marks)
3.
The point $P$ lies on the branch $x>0$ of the hyperbola $\dfrac{x^2}{16}-\dfrac{y^2}{9}=1$ and its distance from the focus $S(5,0)$ is $11$ . Find the coordinates of the two possible positions of $P$ .
(6)
(Total for Question 3 is 6 marks)
4.
Prove directly from the definition that the locus of points equidistant from the point $(a,0)$ and the line $x=-a$ is the parabola $y^2=4ax$ .
(5)
(Total for Question 4 is 5 marks)
5.
An ellipse has its major axis along the $y$ -axis, passes through $(0,\pm13)$ and has eccentricity $\dfrac{5}{13}$ . Find its Cartesian equation and the coordinates of its foci.
(6)
(Total for Question 5 is 6 marks)

FP1-4.3 · Tangents and normals to these curves.

Explanation

Tangents to a parametrised conic are found by implicit or parametric differentiation and then written in a standard form worth memorising. For $y^2=4ax$ at $\left(at^2,2at\right)$ the tangent is $ty=x+at^2$ and the normal is $y+tx=2at+at^3$ .
For $xy=c^2$ at $\left(ct,\dfrac{c}{t}\right)$ the tangent is $x+t^2y=2ct$ and the normal is $t^3x-ty=c\left(t^4-1\right)$ .
For the ellipse at $\left(a\cos\theta,b\sin\theta\right)$ the tangent is $\dfrac{x\cos\theta}{a}+\dfrac{y\sin\theta}{b}=1$ , and for the hyperbola at $\left(a\sec\theta,b\tan\theta\right)$ it is $\dfrac{x\sec\theta}{a}-\dfrac{y\tan\theta}{b}=1$ .
You are also expected to know the tangency conditions: $y=mx+c$ touches $y^2=4ax$ when $c=\dfrac{a}{m}$ , touches $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ when $c^2=a^2m^2+b^2$ , and touches $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$ when $c^2=a^2m^2-b^2$ ; for the rectangular hyperbola $xy=c^2$ the line $y=mx+c_1$ touches it when $c_1^2+4mc^2=0$ , which can only happen when $m<0$ .
All four are expected to be known, and all four come from setting the discriminant of the resulting quadratic to zero, which is also how you find tangents through an external point.

Worked example

Find the equation of the tangent to $y^2=4ax$ at the point $P\left(at^2,2at\right)$ .

1.Differentiating $y^2=4ax$ implicitly gives $2y\dfrac{\mathrm{d}y}{\mathrm{d}x}=4a$ , so $\dfrac{\mathrm{d}y}{\mathrm{d}x}=\dfrac{2a}{y}$ .
2.At $P$ the gradient is $\dfrac{2a}{2at}=\dfrac1t$ .
3.The tangent is $y-2at=\dfrac1t\left(x-at^2\right)$ , that is $ty-2at^2=x-at^2$ .
4.Hence $ty=x+at^2$ .

Answer: $ty=x+at^2$ .

Common mistakes

Don't fall into the trap of using the gradient $\dfrac{\mathrm{d}y}{\mathrm{d}t}$ instead of $\dfrac{\mathrm{d}y}{\mathrm{d}t}\div\dfrac{\mathrm{d}x}{\mathrm{d}t}$ for a parametric curve.
Don't make this mistake: Reflexively writing the normal gradient as $-t$ ; that is only correct when the tangent gradient is $\dfrac1t$ , as it is for the parabola — for $xy=c^2$ the tangent gradient is $-\dfrac1{t^2}$ and the normal gradient is $t^2$ .
Don't fall into the trap of dropping the case $m=0$ when using $c=\dfrac{a}{m}$ ; a horizontal line is never a tangent to $y^2=4ax$ .

Exam tip

Quote the standard tangent form, then verify it by checking that the point itself satisfies it; that single substitution catches almost every algebraic slip.

Tier 1 · Easy

1.
Find the equation of the tangent to the parabola $y^2=12x$ at the point $(12,12)$ .
(3)
(Total for Question 1 is 3 marks)
2.
Find the equation of the tangent to the rectangular hyperbola $xy=16$ at the point $(8,2)$ .
(3)
(Total for Question 2 is 3 marks)

Tier 2 · Standard

1.
Show that the normal to the parabola $y^2=4ax$ at the point $\left(at^2,2at\right)$ has equation $y+tx=2at+at^3$ .
(4)
(Total for Question 1 is 4 marks)
2.
Find the equation of the tangent to the ellipse $\dfrac{x^2}{25}+\dfrac{y^2}{9}=1$ at the point where $\theta=\dfrac{\pi}{3}$ in the parametrisation $\left(5\cos\theta,3\sin\theta\right)$ .
(5)
(Total for Question 2 is 5 marks)
3.
Show that the tangent to the hyperbola $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$ at the point $\left(a\sec\theta,b\tan\theta\right)$ has equation $\dfrac{x\sec\theta}{a}-\dfrac{y\tan\theta}{b}=1$ .
(5)
(Total for Question 3 is 5 marks)

Tier 3 · Hard

1.
Show that the line $y=mx+k$ , with $m\ne0$ , is a tangent to the parabola $y^2=4ax$ if and only if $k=\dfrac{a}{m}$ .
(5)
(Total for Question 1 is 5 marks)
2.
Show that the normal to the rectangular hyperbola $xy=c^2$ at the point $\left(ct,\dfrac{c}{t}\right)$ has equation $t^3x-ty=c\left(t^4-1\right)$ .
(6)
(Total for Question 2 is 6 marks)
3.
Find the equations of the two tangents to the ellipse $\dfrac{x^2}{9}+\dfrac{y^2}{4}=1$ that are parallel to the line $y=2x$ .
(6)
(Total for Question 3 is 6 marks)
4.
Find the equations of the two tangents from the point $(-2,0)$ to the parabola $y^2=8x$ .
(6)
(Total for Question 4 is 6 marks)
5.
The tangent to $y^2=4ax$ at $P\left(at^2,2at\right)$ , with $t\ne0$ , meets the directrix $x=-a$ at the point $Q$ . Show that $PQ$ subtends a right angle at the focus $S(a,0)$ .
(7)
(Total for Question 5 is 7 marks)

FP1-4.4 · Loci problems.

Explanation

A locus question asks for the equation of the curve traced by a moving point, and the reliable method is always the same three steps. First, write the coordinates of the moving point in terms of a parameter, usually $t$ or $\theta$ from the parametrisation of the underlying conic.
Second, set $x$ and $y$ equal to those two expressions. Third, eliminate the parameter between them to leave a relation in $x$ and $y$ alone.
Where a geometric condition is imposed on a chord, first turn it into an algebraic relation between the two parameters, such as $pq=-1$ for a focal chord of $y^2=4ax$ or $pq=-4$ for a chord subtending a right angle at the vertex.
Symmetric functions help: if only $p+q$ and $pq$ appear, use $p^2+q^2=(p+q)^2-2pq$ .
Finally, state any restriction on the range of the locus that the geometry forces, since a bare equation may describe more of a curve than is actually traced.

Worked example

The point $P$ moves on the parabola $y^2=4ax$ and $O$ is the origin. Find the locus of the midpoint $M$ of $OP$ .

1.Write $P$ as $\left(at^2,2at\right)$ , so $M=\left(\dfrac{at^2}{2},\,at\right)$ .
2.Set $x=\dfrac{at^2}{2}$ and $y=at$ , so $t=\dfrac{y}{a}$ .
3.Substituting gives $x=\dfrac{a}{2}\cdot\dfrac{y^2}{a^2}=\dfrac{y^2}{2a}$ .

Answer: The locus is the parabola $y^2=2ax$ .

Common mistakes

Don't fall into the trap of eliminating the parameter from only one of the two coordinate equations.
Don't fall into the trap of forgetting to convert a geometric condition into a relation between the parameters before eliminating them.
Don't fall into the trap of quoting the whole curve when the geometry restricts the locus to part of it.

Exam tip

Label the moving point's coordinates $x$ and $y$ immediately; keeping $\dfrac{at^2}{2}$ in the working to the last line makes it easy to forget to eliminate $t$ .

Tier 1 · Easy

1.
The point $P$ moves on the parabola $y^2=4ax$ and $O$ is the origin. Find the equation of the locus of the midpoint of $OP$ .
(4)
(Total for Question 1 is 4 marks)
2.
The point $P$ moves on the rectangular hyperbola $xy=c^2$ and $O$ is the origin. Find the equation of the locus of the midpoint of $OP$ .
(4)
(Total for Question 2 is 4 marks)

Tier 2 · Standard

1.
The chord $PQ$ of the parabola $y^2=4ax$ subtends a right angle at the vertex $O$ . Show that $PQ$ passes through the fixed point $(4a,0)$ .
(6)
(Total for Question 1 is 6 marks)
2.
Find the equation of the locus of the midpoints of all focal chords of the parabola $y^2=4ax$ .
(6)
(Total for Question 2 is 6 marks)
3.
The point $P$ moves on the ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ and $N$ is the foot of the perpendicular from $P$ to the $x$ -axis. Find the equation of the locus of the midpoint of $PN$ .
(5)
(Total for Question 3 is 5 marks)

Tier 3 · Hard

1.
The tangent to $y^2=4ax$ at $P\left(at^2,2at\right)$ , with $t\ne0$ , meets the $x$ -axis at $T$ and the $y$ -axis at $Y$ . Find the equation of the locus of the midpoint of $TY$ .
(6)
(Total for Question 1 is 6 marks)
2.
The tangent to the rectangular hyperbola $xy=c^2$ at $P\left(ct,\dfrac{c}{t}\right)$ meets the coordinate axes at $A$ and $B$ . Show that $P$ is the midpoint of $AB$ and that the area of triangle $OAB$ is independent of $t$ .
(6)
(Total for Question 2 is 6 marks)
3.
Show that the foot of the perpendicular from the focus $S(a,0)$ to the tangent to $y^2=4ax$ at $P\left(at^2,2at\right)$ always lies on the line $x=0$ .
(6)
(Total for Question 3 is 6 marks)
4.
A variable chord of the rectangular hyperbola $xy=c^2$ has gradient $-1$ . Find the equation of the locus of its midpoint, and state the restriction on the locus.
(7)
(Total for Question 4 is 7 marks)
5.
The normal to $y^2=4ax$ at $P\left(at^2,2at\right)$ meets the $x$ -axis at $G$ . Find the equation of the locus of the midpoint of $PG$ , and identify the curve.
(7)
(Total for Question 5 is 7 marks)

Answer key

Answers begin on a new printed page so the question pack can be completed without the solutions alongside it.

FP1-4.1 · Cartesian and parametric equations for the parabola and rectangular hyperbola, ellipse and hyperbola.

Tier 1 · Easy

Mark scheme for FP1-4.1 Tier 1 · Easy
Question	Scheme	Marks
1	$t=\dfrac{y}{10}$ , so $x=5\left(\dfrac{y}{10}\right)^2=\dfrac{y^2}{20}$ $y^2=20x$ $4a=20$ , so $a=5$	3
(3 marks)		3
Notes
Eliminate $t$ from the linear equation and substitute into the quadratic one. Independent check: the standard parametrisation of $y^2=4ax$ is $\left(at^2,2at\right)$ , and with $a=5$ this is exactly $\left(5t^2,10t\right)$ .
2	$\cos t=\dfrac{x}{6}$ and $\sin t=\dfrac{y}{4}$ $\dfrac{x^2}{36}+\dfrac{y^2}{16}=1$ The semi-axes have lengths $6$ and $4$	3
(3 marks)		3
Notes
Substituting into $\cos^2t+\sin^2t=1$ gives the ellipse directly. Independent check: at $t=0$ the point is $(6,0)$ and at $t=\tfrac{\pi}{2}$ it is $(0,4)$ , which are the ends of the two axes.

Tier 2 · Standard

Mark scheme for FP1-4.1 Tier 2 · Standard
Question	Scheme	Marks
1	$x(x-6)=16$ , so $x^2-6x-16=0$ $(x-8)(x+2)=0$ , giving $x=8$ or $x=-2$ The points are $(8,2)$ and $(-2,-8)$ $4t=8$ gives $t=2$ , and $4t=-2$ gives $t=-\dfrac12$	5
(5 marks)		5
Notes
Substituting the line into the curve gives a quadratic in $x$ with roots $8$ and $-2$ ; the corresponding $y$ values follow from $y=x-6$ . Independent check: $\left(4t,\tfrac4t\right)$ with $t=2$ is $(8,2)$ and with $t=-\tfrac12$ is $(-2,-8)$ , and both satisfy $xy=16$ .
2	$\cos t=\dfrac{x-2}{3}$ and $\sin t=\dfrac{y+1}{5}$ $\dfrac{(x-2)^2}{9}+\dfrac{(y+1)^2}{25}=1$ The centre is $(2,-1)$	4
(4 marks)		4
Notes
The translation shifts the standard ellipse without changing its shape, so the same Pythagorean identity applies to the shifted coordinates. Independent check: at $t=\tfrac{\pi}{2}$ the point is $(2,4)$ , and $\dfrac{0}{9}+\dfrac{25}{25}=1$ as required.
3	$\cosh t=\dfrac{x}{3}$ and $\sinh t=\dfrac{y}{4}$ $\cosh^2t-\sinh^2t=1$ gives $\dfrac{x^2}{9}-\dfrac{y^2}{16}=1$ The asymptotes are $y=\pm\dfrac43x$ Since $\cosh t\ge1$ , only the branch with $x\ge3$ is traced	5
(5 marks)		5
Notes
Eliminating $t$ with the hyperbolic identity gives the standard form with $a=3$ and $b=4$ ; the asymptotes of $\tfrac{x^2}{a^2}-\tfrac{y^2}{b^2}=1$ are $y=\pm\tfrac{b}{a}x$ . Independent check: at $t=0$ the point is $(3,0)$ , the right-hand vertex, and $x=3\cosh t$ can never be negative.

Tier 3 · Hard

Mark scheme for FP1-4.1 Tier 3 · Hard
Question	Scheme	Marks
1	Gradient $=\dfrac{2ap-2aq}{ap^2-aq^2}=\dfrac{2}{p+q}$ $y-2ap=\dfrac{2}{p+q}\left(x-ap^2\right)$ $(p+q)y-2ap(p+q)=2x-2ap^2$ $(p+q)y=2x+2apq$	5
(5 marks)		5
Notes
The difference of the $y$ -coordinates is $2a(p-q)$ and of the $x$ -coordinates is $a(p-q)(p+q)$ , so the gradient simplifies to $\tfrac{2}{p+q}$ . Expanding the point-gradient form and cancelling $2ap^2$ leaves the stated equation. Independent check with $a=1$ , $p=1$ , $q=3$ : the points are $(1,2)$ and $(9,6)$ , and $4y=2x+6$ passes through both.
2	Gradient $=\dfrac{c/p-c/q}{cp-cq}=\dfrac{c(q-p)/(pq)}{c(p-q)}=-\dfrac{1}{pq}$ $y-\dfrac{c}{p}=-\dfrac{1}{pq}\left(x-cp\right)$ Multiplying by $pq$ gives $pqy-cq=-x+cp$ $x+pqy=c(p+q)$	5
(5 marks)		5
Notes
Simplify the gradient by writing the numerator over $pq$ ; the factor $q-p$ cancels against $p-q$ leaving $-\tfrac1{pq}$ . Clearing the fraction in the point-gradient form gives the required chord. Independent check with $c=2$ , $p=1$ , $q=4$ : the points are $(2,2)$ and $(8,\tfrac12)$ , and $x+4y=10$ passes through both.
3	Substituting gives $b^2x^2+a^2(mx+k)^2=a^2b^2$ $\left(b^2+a^2m^2\right)x^2+2a^2mkx+a^2\left(k^2-b^2\right)=0$ Discriminant $=4a^4m^2k^2-4a^2\left(b^2+a^2m^2\right)\left(k^2-b^2\right)$ $=4a^2b^2\left(a^2m^2+b^2-k^2\right)$ Two distinct points require this to be positive, that is $k^2<a^2m^2+b^2$	6
(6 marks)		6
Notes
Substituting the line gives a quadratic in $x$ ; expanding the discriminant and cancelling the $a^4m^2k^2$ terms leaves $4a^2b^2\left(a^2m^2+b^2-k^2\right)$ . Independent check: for the circle $a=b=1$ the condition becomes $k^2<m^2+1$ , that is $\dfrac{\|k\|}{\sqrt{1+m^2}}<1$ , which is exactly the statement that the perpendicular distance from the centre to the line is less than the radius.
4	$\cosh t=\dfrac{x-3}{2}$ and $\sinh t=\dfrac{y+1}{5}$ $\dfrac{(x-3)^2}{4}-\dfrac{(y+1)^2}{25}=1$ The centre is $(3,-1)$ The asymptotes are $y+1=\pm\dfrac52(x-3)$ Since $\cosh t\ge1$ , the parametrisation gives $x\ge5$ only	6
(6 marks)		6
Notes
Apply the hyperbolic identity to the translated coordinates; the asymptotes of the translated hyperbola pass through its centre with gradients $\pm\tfrac{b}{a}=\pm\tfrac52$ . Independent check: at $t=0$ the point is $(5,-1)$ , the vertex of the right-hand branch, and $\dfrac{4}{4}-0=1$ as required.
5	The asymptotes give $\dfrac{b}{a}=\dfrac34$ , so $b^2=\dfrac{9a^2}{16}$ Substituting the point: $\dfrac{64}{a^2}-\dfrac{27}{9a^2/16}=1$ $\dfrac{64}{a^2}-\dfrac{48}{a^2}=1$ , so $\dfrac{16}{a^2}=1$ and $a^2=16$ $b^2=9$ $\dfrac{x^2}{16}-\dfrac{y^2}{9}=1$	6
(6 marks)		6
Notes
The asymptote gradients fix the ratio $\tfrac{b}{a}$ , and substituting the given point then determines $a^2$ . Independent check: $\dfrac{64}{16}-\dfrac{27}{9}=4-3=1$ , so the point does lie on the curve, and the asymptote gradients are $\pm\tfrac34$ as required.

FP1-4.2 · The focus-directrix properties of the parabola, ellipse and hyperbola, including the eccentricity.

Tier 1 · Easy

Mark scheme for FP1-4.2 Tier 1 · Easy
Question	Scheme	Marks
1	$4a=16$ , so $a=4$ The focus is $(4,0)$ and the directrix is $x=-4$	2
(2 marks)		2
Notes
Compare with the standard form $y^2=4ax$ to get $a=4$ , then quote the focus $(a,0)$ and the directrix $x=-a$ . Independent check: the point $(4,8)$ lies on the curve, its distance from $(4,0)$ is $8$ , and its distance from $x=-4$ is $4+4=8$ , so the focus-directrix property holds with $e=1$ .
2	$16=25\left(1-e^2\right)$ , so $e^2=\dfrac9{25}$ $e=\dfrac35$ The foci are $(\pm3,0)$	3
(3 marks)		3
Notes
Substitute $a^2=25$ and $b^2=16$ into $b^2=a^2\left(1-e^2\right)$ , then use $(\pm ae,0)$ . Independent check: the point $(0,4)$ is $\sqrt{9+16}=5=a$ from each focus, and the two focal distances sum to $10=2a$ as required.

Tier 2 · Standard

Mark scheme for FP1-4.2 Tier 2 · Standard
Question	Scheme	Marks
1	$16=9\left(e^2-1\right)$ , so $e^2=\dfrac{25}{9}$ $e=\dfrac53$ The foci are $(\pm5,0)$ The directrices are $x=\pm\dfrac95$	4
(4 marks)		4
Notes
Use $b^2=a^2\left(e^2-1\right)$ with $a^2=9$ and $b^2=16$ , then $(\pm ae,0)$ and $x=\pm\tfrac{a}{e}$ . Independent check: the vertex $(3,0)$ has focal distance $5-3=2$ and directrix distance $3-\tfrac95=\tfrac65$ , and $\tfrac53\times\tfrac65=2$ , confirming the focus-directrix property.
2	$ae=4$ and $e=\dfrac23$ give $a=6$ $b^2=36\left(1-\dfrac49\right)=20$ $\dfrac{x^2}{36}+\dfrac{y^2}{20}=1$	4
(4 marks)		4
Notes
The focus coordinate gives $a$ directly, and $b^2=a^2\left(1-e^2\right)$ then gives $b^2=36\times\tfrac59=20$ . Independent check: with $a^2=36$ and $b^2=20$ , $1-e^2=\tfrac{20}{36}=\tfrac59$ , so $e^2=\tfrac49$ and $e=\tfrac23$ as required.
3	$4a=12$ , so $a=3$ and the directrix is $x=-3$ The focal distance equals the directrix distance, so $x+3=7$ and $x=4$ $y^2=48$ , so $y=\pm4\sqrt3$ $P$ is $\left(4,4\sqrt3\right)$ or $\left(4,-4\sqrt3\right)$	5
(5 marks)		5
Notes
For a parabola $e=1$ , so the focal distance equals the perpendicular distance to the directrix, namely $x+a$ . Setting $x+3=7$ gives $x=4$ and hence $y^2=12\times4=48$ . Independent check: the distance from $\left(4,4\sqrt3\right)$ to $(3,0)$ is $\sqrt{1+48}=7$ .

Tier 3 · Hard

Mark scheme for FP1-4.2 Tier 3 · Hard
Question	Scheme	Marks
1	$SP=e\left(\dfrac{a}{e}-x\right)=a-ex$ $S'P=e\left(x+\dfrac{a}{e}\right)=a+ex$ Adding gives $SP+S'P=2a$ The result is independent of the position of $P$	6
(6 marks)		6
Notes
Each focal distance is $e$ times the perpendicular distance to the corresponding directrix $x=\pm\tfrac{a}{e}$ ; the $ex$ terms cancel on addition. Independent check with $a=5$ , $b=4$ , $e=\tfrac35$ at $P(0,4)$ : both focal distances are $\sqrt{9+16}=5$ and their sum is $10=2a$ ; at $P(5,0)$ they are $2$ and $8$ , again summing to $10$ .
2	At the focus $x=ae$ , so $\dfrac{a^2e^2}{a^2}+\dfrac{y^2}{b^2}=1$ $y^2=b^2\left(1-e^2\right)$ $b^2=a^2\left(1-e^2\right)$ gives $1-e^2=\dfrac{b^2}{a^2}$ $y^2=\dfrac{b^4}{a^2}$ , so $y=\pm\dfrac{b^2}{a}$ and the chord has length $\dfrac{2b^2}{a}$	5
(5 marks)		5
Notes
Substituting $x=ae$ into the ellipse and then eliminating $1-e^2$ using $b^2=a^2\left(1-e^2\right)$ gives $y=\pm\tfrac{b^2}{a}$ . Independent check with $a=5$ , $b=4$ : the focus is $(3,0)$ , and substituting $x=3$ gives $\tfrac{9}{25}+\tfrac{y^2}{16}=1$ , so $y=\pm\tfrac{16}{5}$ and the length is $\tfrac{32}{5}=\tfrac{2\times16}{5}$ .
3	$9=16\left(e^2-1\right)$ gives $e=\dfrac54$ On this branch $SP=ex-a=\dfrac54x-4$ $\dfrac54x-4=11$ gives $x=12$ $y^2=9\left(\dfrac{144}{16}-1\right)=72$ , so $y=\pm6\sqrt2$ $P$ is $\left(12,6\sqrt2\right)$ or $\left(12,-6\sqrt2\right)$	6
(6 marks)		6
Notes
Compute the eccentricity from $b^2=a^2\left(e^2-1\right)$ , use the focal-distance formula $SP=ex-a$ valid on the branch nearer $S$ , and then substitute back into the hyperbola. Independent check: the distance from $\left(12,6\sqrt2\right)$ to $(5,0)$ is $\sqrt{49+72}=\sqrt{121}=11$ .
4	$SP^2=(x-a)^2+y^2$ and the directrix distance is $\|x+a\|$ Equating squares: $(x-a)^2+y^2=(x+a)^2$ $x^2-2ax+a^2+y^2=x^2+2ax+a^2$ $y^2=4ax$	5
(5 marks)		5
Notes
Write both distances, square to remove the modulus and the square root, and cancel the common terms; only $-2ax$ and $+2ax$ survive. Independent check: for $a=3$ the locus is $y^2=12x$ , and the point $(3,6)$ on it is $6$ from $(3,0)$ and $6$ from the line $x=-3$ .
5	The semi-major axis is $13$ and lies along the $y$ -axis The semi-minor axis satisfies $b^2=169\left(1-\dfrac{25}{169}\right)=144$ $\dfrac{x^2}{144}+\dfrac{y^2}{169}=1$ The foci lie on the $y$ -axis at $(0,\pm ae)=(0,\pm5)$	6
(6 marks)		6
Notes
When the major axis is vertical, the roles of the two denominators are exchanged: the larger denominator sits under $y^2$ , and the foci are on the $y$ -axis. With $a=13$ and $e=\tfrac5{13}$ , $b^2=169-25=144$ . Independent check: the point $(12,0)$ lies on the curve and its distances to $(0,5)$ and $(0,-5)$ are both $13$ , summing to $26=2a$ .

FP1-4.3 · Tangents and normals to these curves.

Tier 1 · Easy

Mark scheme for FP1-4.3 Tier 1 · Easy
Question	Scheme	Marks
1	$4a=12$ gives $a=3$ , and $\left(at^2,2at\right)=(12,12)$ gives $t=2$ The tangent $ty=x+at^2$ becomes $2y=x+12$	3
(3 marks)		3
Notes
Identify $a=3$ , then $2at=12$ gives $t=2$ and $at^2=12$ confirms it. Substituting into $ty=x+at^2$ gives $2y=x+12$ . Independent check by implicit differentiation: $2y\tfrac{\mathrm{d}y}{\mathrm{d}x}=12$ , so at $(12,12)$ the gradient is $\tfrac12$ , and $y-12=\tfrac12(x-12)$ rearranges to $2y=x+12$ .
2	$c=4$ and $\left(ct,\tfrac{c}{t}\right)=(8,2)$ gives $t=2$ The tangent $x+t^2y=2ct$ becomes $x+4y=16$	3
(3 marks)		3
Notes
With $c^2=16$ , $c=4$ and $4t=8$ gives $t=2$ . Substituting into $x+t^2y=2ct$ gives $x+4y=16$ . Independent check: $y=16x^{-1}$ gives $\tfrac{\mathrm{d}y}{\mathrm{d}x}=-16x^{-2}=-\tfrac14$ at $x=8$ , and $y-2=-\tfrac14(x-8)$ rearranges to $x+4y=16$ .

Tier 2 · Standard

Mark scheme for FP1-4.3 Tier 2 · Standard
Question	Scheme	Marks
1	The tangent gradient is $\dfrac1t$ , so the normal gradient is $-t$ $y-2at=-t\left(x-at^2\right)$ $y-2at=-tx+at^3$ $y+tx=2at+at^3$	4
(4 marks)		4
Notes
Parametric differentiation gives $\dfrac{\mathrm{d}y}{\mathrm{d}x}=\dfrac{2a}{2at}=\dfrac1t$ , so the normal gradient is $-t$ ; substituting into the point-gradient form and rearranging gives the result. Independent check with $a=1$ , $t=2$ : the point is $(4,4)$ and the normal is $y+2x=12$ , which passes through $(4,4)$ and is perpendicular to the tangent $2y=x+4$ of gradient $\tfrac12$ .
2	The tangent is $\dfrac{x\cos\theta}{5}+\dfrac{y\sin\theta}{3}=1$ $\cos\tfrac{\pi}{3}=\tfrac12$ and $\sin\tfrac{\pi}{3}=\tfrac{\sqrt3}{2}$ $\dfrac{x}{10}+\dfrac{y\sqrt3}{6}=1$ $3x+5\sqrt3y=30$	5
(5 marks)		5
Notes
Substitute the parameter values into the standard tangent form and clear denominators by multiplying by $30$ . Independent check: the point of contact is $\left(\tfrac52,\tfrac{3\sqrt3}{2}\right)$ , and $3\left(\tfrac52\right)+5\sqrt3\left(\tfrac{3\sqrt3}{2}\right)=\tfrac{15}{2}+\tfrac{45}{2}=30$ .
3	$\dfrac{\mathrm{d}x}{\mathrm{d}\theta}=a\sec\theta\tan\theta$ and $\dfrac{\mathrm{d}y}{\mathrm{d}\theta}=b\sec^2\theta$ $\dfrac{\mathrm{d}y}{\mathrm{d}x}=\dfrac{b\sec\theta}{a\tan\theta}$ $y-b\tan\theta=\dfrac{b\sec\theta}{a\tan\theta}\left(x-a\sec\theta\right)$ Multiplying by $\dfrac{\tan\theta}{b}$ and using $\sec^2\theta-\tan^2\theta=1$ gives $\dfrac{x\sec\theta}{a}-\dfrac{y\tan\theta}{b}=1$	5
(5 marks)		5
Notes
Parametric differentiation gives the gradient $\dfrac{b\sec\theta}{a\tan\theta}$ ; rearranging the point-gradient equation produces $\dfrac{x\sec\theta}{a}-\dfrac{y\tan\theta}{b}=\sec^2\theta-\tan^2\theta$ , which is $1$ . Independent check with $a=4$ , $b=3$ , $\theta=\tfrac{\pi}{4}$ : the point is $\left(4\sqrt2,3\right)$ and the tangent is $\tfrac{x\sqrt2}{4}-\tfrac{y}{3}=1$ , which the point satisfies since $\tfrac{4\sqrt2\cdot\sqrt2}{4}-1=2-1=1$ .

Tier 3 · Hard

Mark scheme for FP1-4.3 Tier 3 · Hard
Question	Scheme	Marks
1	Substituting gives $(mx+k)^2=4ax$ $m^2x^2+2\left(mk-2a\right)x+k^2=0$ Tangency requires the discriminant to vanish: $4\left(mk-2a\right)^2-4m^2k^2=0$ $-4amk+4a^2=0$ , so $k=\dfrac{a}{m}$	5
(5 marks)		5
Notes
Expanding the discriminant gives $m^2k^2-4amk+4a^2-m^2k^2=4a\left(a-mk\right)$ , which vanishes exactly when $mk=a$ . Independent check with $a=1$ and $m=2$ : the line $y=2x+\tfrac12$ meets $y^2=4x$ where $4x^2+2x+\tfrac14-4x=0$ , that is $\left(2x-\tfrac12\right)^2=0$ , a repeated root at $x=\tfrac14$ .
2	$y=\dfrac{c^2}{x}$ gives $\dfrac{\mathrm{d}y}{\mathrm{d}x}=-\dfrac{c^2}{x^2}=-\dfrac1{t^2}$ at the point The normal gradient is $t^2$ $y-\dfrac{c}{t}=t^2\left(x-ct\right)$ Multiplying by $t$ gives $ty-c=t^3x-ct^4$ $t^3x-ty=c\left(t^4-1\right)$	6
(6 marks)		6
Notes
Differentiate the curve, evaluate at $x=ct$ to get the tangent gradient $-t^{-2}$ , and take the negative reciprocal $t^2$ for the normal; clearing the fraction gives the stated form. Independent check with $c=1$ , $t=2$ : the point is $\left(2,\tfrac12\right)$ and the normal is $8x-2y=15$ , which the point satisfies since $16-1=15$ .
3	A tangent of gradient $m$ satisfies $k^2=a^2m^2+b^2$ With $a^2=9$ , $b^2=4$ and $m=2$ : $k^2=36+4=40$ $k=\pm2\sqrt{10}$ $y=2x+2\sqrt{10}$ and $y=2x-2\sqrt{10}$	6
(6 marks)		6
Notes
Use the tangency condition $k^2=a^2m^2+b^2$ , which comes from setting the discriminant of the substituted quadratic to zero. Independent check: substituting $y=2x+2\sqrt{10}$ into the ellipse gives $4x^2+9\left(2x+2\sqrt{10}\right)^2=36$ , that is $40x^2+72\sqrt{10}x+324=0$ , whose discriminant is $51840-51840=0$ , confirming a repeated root.
4	$4a=8$ gives $a=2$ , so a tangent has the form $y=mx+\dfrac2m$ Through $(-2,0)$ : $0=-2m+\dfrac2m$ $2m^2=2$ , so $m=\pm1$ $y=x+2$ and $y=-x-2$	6
(6 marks)		6
Notes
Every non-horizontal tangent is $y=mx+\tfrac{a}{m}$ ; imposing that it passes through the external point gives a quadratic in $m$ with roots $\pm1$ . Independent check: $y=x+2$ meets $y^2=8x$ where $(x+2)^2=8x$ , that is $(x-2)^2=0$ , a repeated root at $x=2$ ; the point $(-2,0)$ lies on the directrix $x=-2$ , and the two tangents from any directrix point are indeed perpendicular, as $1\times(-1)=-1$ confirms.
5	The tangent is $ty=x+at^2$ , so at $x=-a$ , $y=\dfrac{a\left(t^2-1\right)}{t}$ $Q$ is $\left(-a,\dfrac{a\left(t^2-1\right)}{t}\right)$ $\overrightarrow{SP}=a\left(t^2-1,\ 2t\right)$ and $\overrightarrow{SQ}=\left(-2a,\ \dfrac{a\left(t^2-1\right)}{t}\right)$ $\overrightarrow{SP}\cdot\overrightarrow{SQ}=-2a^2\left(t^2-1\right)+2a^2\left(t^2-1\right)=0$ Hence $\angle PSQ=90^\circ$	7
(7 marks)		7
Notes
Substituting $x=-a$ into the tangent gives the $y$ -coordinate of $Q$ ; forming the two displacement vectors from the focus and taking their scalar product gives exactly cancelling terms. Independent check with $a=1$ , $t=2$ : $P(4,4)$ , $Q\left(-1,\tfrac32\right)$ and $S(1,0)$ , so $\overrightarrow{SP}=(3,4)$ and $\overrightarrow{SQ}=\left(-2,\tfrac32\right)$ , whose scalar product is $-6+6=0$ .

FP1-4.4 · Loci problems.

Tier 1 · Easy

Mark scheme for FP1-4.4 Tier 1 · Easy
Question	Scheme	Marks
1	$P=\left(at^2,2at\right)$ , so the midpoint is $\left(\dfrac{at^2}{2},at\right)$ $x=\dfrac{at^2}{2}$ and $y=at$ , so $t=\dfrac{y}{a}$ $x=\dfrac{y^2}{2a}$ , that is $y^2=2ax$	4
(4 marks)		4
Notes
Parametrise $P$ , halve both coordinates and eliminate $t$ . Independent check with $a=1$ and $t=2$ : $P(4,4)$ , midpoint $(2,2)$ , and $2^2=2\times1\times2=4$ .
2	$P=\left(ct,\dfrac{c}{t}\right)$ , so the midpoint is $\left(\dfrac{ct}{2},\dfrac{c}{2t}\right)$ Multiplying the coordinates gives $xy=\dfrac{c^2}{4}$	4
(4 marks)		4
Notes
The product of the halved coordinates is a quarter of the original product, so the locus is another rectangular hyperbola. Independent check with $c=2$ and $t=1$ : $P(2,2)$ , midpoint $(1,1)$ , and $1\times1=\tfrac{4}{4}=1$ .

Tier 2 · Standard

Mark scheme for FP1-4.4 Tier 2 · Standard
Question	Scheme	Marks
1	$\overrightarrow{OP}\cdot\overrightarrow{OQ}=a^2p^2q^2+4a^2pq=0$ $pq\left(pq+4\right)=0$ and $pq\ne0$ , so $pq=-4$ The chord is $(p+q)y=2x+2apq$ Substituting $pq=-4$ gives $(p+q)y=2x-8a$ This meets $y=0$ where $x=4a$ , independently of $p$ and $q$	6
(6 marks)		6
Notes
The right-angle condition makes the scalar product of the position vectors zero, which forces $pq=-4$ ; substituting into the standard chord equation shows the $x$ -intercept is always $4a$ . Independent check with $a=1$ , $p=2$ , $q=-2$ : $P(4,4)$ and $Q(4,-4)$ satisfy $\overrightarrow{OP}\cdot\overrightarrow{OQ}=16-16=0$ , and the chord $x=4$ passes through $(4,0)$ .
2	A focal chord has $pq=-1$ The midpoint is $\left(\dfrac{a\left(p^2+q^2\right)}{2},\ a(p+q)\right)$ With $s=p+q$ , $p^2+q^2=s^2+2$ , so $x=\dfrac{a\left(s^2+2\right)}{2}$ and $y=as$ Eliminating $s=\dfrac{y}{a}$ gives $x=\dfrac{y^2}{2a}+a$ $y^2=2a(x-a)$	6
(6 marks)		6
Notes
Use the focal-chord condition $pq=-1$ and the symmetric identity $p^2+q^2=(p+q)^2-2pq$ to express both coordinates in terms of $s=p+q$ alone, then eliminate $s$ . Independent check with $a=1$ , $p=1$ , $q=-1$ : the chord joins $(1,2)$ and $(1,-2)$ , its midpoint is $(1,0)$ , and $0=2(1-1)$ .
3	$P=\left(a\cos\theta,b\sin\theta\right)$ and $N=\left(a\cos\theta,0\right)$ The midpoint is $\left(a\cos\theta,\ \dfrac{b\sin\theta}{2}\right)$ $\cos\theta=\dfrac{x}{a}$ and $\sin\theta=\dfrac{2y}{b}$ $\dfrac{x^2}{a^2}+\dfrac{4y^2}{b^2}=1$	5
(5 marks)		5
Notes
The midpoint keeps the $x$ -coordinate and halves the $y$ -coordinate, so the Pythagorean identity gives an ellipse with the same semi-major axis and half the semi-minor axis. Independent check with $a=5$ , $b=4$ at $\theta=\tfrac{\pi}{2}$ : $P(0,4)$ , $N(0,0)$ , midpoint $(0,2)$ , and $0+\tfrac{4\times4}{16}=1$ .

Tier 3 · Hard

Mark scheme for FP1-4.4 Tier 3 · Hard
Question	Scheme	Marks
1	The tangent is $ty=x+at^2$ , so $T=\left(-at^2,0\right)$ and $Y=\left(0,at\right)$ The midpoint is $\left(-\dfrac{at^2}{2},\ \dfrac{at}{2}\right)$ $y=\dfrac{at}{2}$ gives $t=\dfrac{2y}{a}$ $x=-\dfrac{a}{2}\cdot\dfrac{4y^2}{a^2}=-\dfrac{2y^2}{a}$ $y^2=-\dfrac{ax}{2}$	6
(6 marks)		6
Notes
Setting $y=0$ and then $x=0$ in the tangent gives the two intercepts; halving their sum and eliminating $t$ gives a parabola opening in the negative $x$ direction. Independent check with $a=1$ , $t=2$ : the tangent $2y=x+4$ meets the axes at $(-4,0)$ and $(0,2)$ , whose midpoint is $(-2,1)$ , and $1=-\tfrac{-2}{2}=1$ .
2	The tangent is $x+t^2y=2ct$ , so $A=(2ct,0)$ and $B=\left(0,\dfrac{2c}{t}\right)$ The midpoint of $AB$ is $\left(ct,\dfrac{c}{t}\right)$ , which is $P$ Area $=\dfrac12\times2ct\times\dfrac{2c}{t}$ $=2c^2$ , independent of $t$	6
(6 marks)		6
Notes
Setting $y=0$ and $x=0$ in the tangent gives the two intercepts; their midpoint is exactly $P$ , and the two factors of $t$ cancel in the area. Independent check with $c=2$ , $t=1$ : the tangent is $x+y=4$ , meeting the axes at $(4,0)$ and $(0,4)$ , whose midpoint is $(2,2)=P$ and whose triangle has area $8=2\times4$ .
3	The tangent is $x-ty+at^2=0$ , with gradient $\dfrac1t$ The perpendicular through $S$ has gradient $-t$ : $y=-t(x-a)$ Substituting into the tangent: $x+t^2(x-a)+at^2=0$ $x\left(1+t^2\right)=0$ , so $x=0$ and $y=at$ The foot is $(0,at)$ , which lies on the tangent at the vertex	6
(6 marks)		6
Notes
Solve the tangent and the perpendicular from the focus simultaneously; the $at^2$ terms cancel, forcing $x=0$ whatever the value of $t$ . Independent check with $a=1$ , $t=2$ : the tangent is $2y=x+4$ and the perpendicular from $(1,0)$ is $y=-2(x-1)$ ; solving gives $x=0$ , $y=2$ , and $(0,2)$ lies on both lines.
4	The chord joining $\left(cp,\tfrac{c}{p}\right)$ and $\left(cq,\tfrac{c}{q}\right)$ has gradient $-\dfrac1{pq}$ Gradient $-1$ gives $pq=1$ The midpoint is $\left(\dfrac{c(p+q)}{2},\ \dfrac{c(p+q)}{2pq}\right)$ With $pq=1$ both coordinates equal $\dfrac{c(p+q)}{2}$ , so the locus is the line $y=x$ Since $p$ and $q$ are distinct and $pq=1$ , $\|p+q\|>2$ , so $\|x\|>c$	7
(7 marks)		7
Notes
The gradient condition fixes $pq=1$ , after which the two midpoint coordinates coincide. The restriction follows because $p+q$ with $pq=1$ satisfies $\|p+q\|\ge2$ , with equality only when $p=q$ , which a genuine chord excludes. Independent check with $c=1$ , $p=2$ , $q=\tfrac12$ : the points are $\left(2,\tfrac12\right)$ and $\left(\tfrac12,2\right)$ , the chord has gradient $-1$ , and the midpoint $\left(\tfrac54,\tfrac54\right)$ lies on $y=x$ with $\tfrac54>1$ .
5	The normal is $y+tx=2at+at^3$ , so at $y=0$ , $x=2a+at^2$ $G=\left(2a+at^2,\ 0\right)$ The midpoint is $\left(\dfrac{at^2+2a+at^2}{2},\ at\right)=\left(at^2+a,\ at\right)$ $t=\dfrac{y}{a}$ gives $x=\dfrac{y^2}{a}+a$ $y^2=a(x-a)$ , a parabola with vertex $(a,0)$ and the same axis	7
(7 marks)		7
Notes
Setting $y=0$ in the normal gives $G$ ; averaging the coordinates of $P$ and $G$ and eliminating $t$ gives a parabola whose vertex sits at the focus of the original curve. Independent check with $a=1$ , $t=2$ : $P(4,4)$ , the normal $y+2x=12$ meets the $x$ -axis at $(6,0)$ , the midpoint is $(5,2)$ , and $2^2=1\times(5-1)=4$ .