Pearson Edexcel IAL Mathematics FP1.4 Coordinate systems Question Bank

Practise parabola and hyperbola coordinate work, including parametric points, focus-directrix geometry, tangents, normals and area conditions.

Exam points

use y^2=4ax, xy=c^2 and parametric points to find coordinates or constants
link focus and directrix facts to exact lengths, perimeters or triangle areas
derive tangent or normal equations and use intersections to determine points

Covered syllabus points

FP1.4.1 - Cartesian equations for the parabola FP1.4.2 - Parametric equations for conics FP1.4.3 - Focus-directrix property of the parabola FP1.4.4 - Tangents and normals to conics

[Maximum number: 13]

The rectangular hyperbola H has equation $x y=c^{2}$ where c is a positive constant.

The point $P\left(c t, \frac{c}{t}\right)$ , where t>0, lies on H

(a)

Use calculus to show that an equation of the normal to H at P is

t^{3} x-t y=c\left(t^{4}-1\right)

The parabola C has equation $y^{2}=6 x$
The normal to H at the point with coordinates (8,2) meets C at the point Q where y>0

[ 4 ]

$xy=c^2\Rightarrow y=c^2x^{-1}$ $\frac{dy}{dx}=-c^2x^{-2}$
At P(ct,c/t), $\frac{dy}{dx}=-\frac{1}{t^2}$ , so the normal gradient is $t^2$ .
$y-\frac ct=t^2(x-ct)$ $ty-c=t^3x-ct^4$ $t^3x-ty=c(t^4-1)$
B1 M1 M1 A1

(b)

Determine the exact coordinates of Q

Given that
- the point R is the focus of C
- the line l is the directrix of C
- the line through Q and R meets l at the point S

[ 4 ]

At (8,2), $c^2=16$ , so c=4, and ct=8, so t=2.
Normal: 8x-2y=4(16-1)=60, so y=4x-30.
With $y^2=6x$ : $(4x-30)^2=6x$ .
$16x^2-246x+900=0$ $x=6\text{ or }\frac{75}{8}$
Since y>0, $Q=\left(\frac{75}{8},\frac{15}{2}\right)$ .
B1 M1 dM1 A1

(c)

determine the exact length of QS

[ 5 ]

For $y^2=6x$ , 4a=6, so focus $R=\left(\frac32,0\right)$ and directrix $x=-\frac32$ .
Line through $Q\left(\frac{75}{8},\frac{15}{2}\right)$ and $R\left(\frac32,0\right)$ has gradient $\frac{20}{21}$ .
$y=\frac{20}{21}\left(x-\frac32\right)$
At $x=-\frac32$ , $y=-\frac{20}{7}$ , so $S=\left(-\frac32,-\frac{20}{7}\right)$ .
$QS=\sqrt{\left(\frac{75}{8}+\frac32\right)^2+\left(\frac{15}{2}+\frac{20}{7}\right)^2}=\frac{2925}{56}$
B1 M1 M1 M1 A1

Source markscheme diagram

[Maximum number: 10]

The parabola C has equation $y^{2}=20 x$

The point P on C has coordinates ( $5 p^{2}, 10 p$ ) where p is a non-zero constant.

(a)

Use calculus to show that the tangent to C at P has equation\n\n $py-x=5p^2$

[ 3 ]

$y^2=20x\Rightarrow 2y\frac{dy}{dx}=20 \Rightarrow \frac{dy}{dx}=\frac{10}{y}.$
At $P(5p^2,10p)$ ,
$\frac{dy}{dx}=\frac{10}{10p}=\frac1p.$

M1

$y-10p=\frac1p(x-5p^2).$

M1

$py-x=5p^2.$

A1*

(b)

The tangent to C at P meets the y-axis at the point A. Write down the coordinates of A.

[ 1 ]

(c)

The point S is the focus of C. Write down the coordinates of S.

[ 1 ]

(d)

The straight line $l_1$ passes through A and S. The straight line $l_2$ passes through O and P, where O is the origin. Given that $l_1$ and $l_2$ intersect at the point B, show that the coordinates of B satisfy the equation\n\n $2x^2+y^2=10x$

[ 5 ]

$l_2:\quad y=\frac{2}{p}x.$

B1

$l_1:\quad y=-p(x-5)$
or
y=-px+5p.

M1

Using $p=\frac{2x}{y}$ ,
$y=-\frac{2x}{y}(x-5).$

M1

$y^2=-2x^2+10x$
so
$2x^2+y^2=10x.$

A1 A1*

All question bank results loaded

Covered syllabus points

FP1.4 - Coordinate systems question 1

Question (a)

Question (b)

Question (c)

FP1.4 - Coordinate systems question 2

Question (a)

Question (b)

Question (c)

Question (d)