AS Level/ A-Level Further Pure 1 – A* Further Mathematics Pearson Edexcel Summary Notes (8FM0) (9FM0)

FURTHER PURE 1


Chapter 1 : Vectors
Vector Product /Cross product) :
Refer back to CPI notes
1

a x
1 =
/9/16/ sino 1 b R
A X



equation of lines :
x

&
-
L
1
"




49
↓
Direction of I determined
by (1 4)
A
↓ is the 1 0
x =
N
-




-
b =
1 -




A

right-handed screw rule. OR 2 Xb =
2 Where AXb =
C o
.




7
i 1 x
1 =
1
4 bxq 0 Direction cosines
-




1 x =
1 x
1 = :
-




-

-


L xj =
-




1
let A = 9 1 +
A21 + As 1 ↓ 6
-
if 1 = x1 +
y +
21
1
,
x
1 = O (C
Ya
=

b b, l (0SX m cosB n cos0
b c1
=

bs1
= = = = =
= + + a I ·
a




a
2
+y
(4


d
+




=
AX1 l + M + D = = =

>Us A 43 A -
:
=
= -



= + I
..
+ M2 +
D = 1
=
(AcDs-AsDc)[ -19 Ds-AsDa)[ ,
+
(9 ,
b. -Acbi) L

l equation of normal (

Areas & volumes


a a asi
. (b x
2) =
4 .
(2x1)
. XC
-




B triple scalar product :
1 =
1




B C 4 .

(1xb) =
0
>

D
D D
A
>
A D 1
C C
C T
7
Parallelogram
A B C A ↓
> >

= Ac AB
-




Area 1 x Area = x
AD parallele piped A B
pyramid B Tetrahedron


FB (FC ( 5 AB (1 x FD) 5 13 ↑D)
>

(1
-




V = . x AD V = .

V = .
x




Point of intersection betweenI T
and :




>
intersects in a line : a vector that is perpendicular to both n ,
and n = lies in both planes
T1 e. g. A ,: .

(= = 2
find point of intersection,


/-
Ha
(1) 5
|e+ 1 211 2y (1) 2 x
3y
Ni =
5
=
1 z +
12 : =
-


= - -
.
,




.
Tz O 4) ...
2x
2y 3 by =
0
- -
. . .
=




nin n ,
x nc =
( = )x(-) (i) =

0 -




0 : -




2y
-

( -




6y) = -

5


5
4y
=
-




Shortest distance between 2 skew lines :
= - ,
x =



: -
xi
r,
= a + xb F = -


A
*

MA
A
P2 2 +

&
=

normal to 1 and 1 = b x


d

d =
(2 q) -
.
((x1)
X

C b x d




Chapter 2 : Conic Sections
PARABOLA

Y loci proof :
y= 496




↑ i
X I X
Pl at2 2at)
If 2 JaJ
,
=
&
1 .

PX = x + A 2 .



y 49)
y
=




AB2 # =
-




-
PA = +
PB PA = ( -


291 +
9 + 494 I =
25a(

( 94
A
s
= (x -


a) +
y2 = + 2ax +
OR 2
2y
= 49
O




m
= x' -

2ax + a +
y PA =
(c) +
a)
=
+ =

2
PA
A
.
=
PX ·

2 y = 2at ; =
20
y 49/
() = -

A then PX = PA' So (1) +
a) " = (11 a) -

+
y
,
I =
at2 ; =
2at
Focus : (a 0 ) , ( + 29x + a = x -


2ax + q +
y2
Directrix : () = -

A
y
" = 49s) Cartesian form - = =
&
: = 4 96 E) DX PA Parabola general point () at 2at Parametric form
y y
= : = =

, RECIPROCAL Rectangular Hyperbola loci proof :




Y At
Y
general point :
<= Ct , y
= = parametric ↓ Plat" sat) , x =
2 y
=
at
t
4(π)
:




y
=
c Cartesian + =
4 .. x =

2
Plct E
M/A y
,

x

py = C Reciprocal ,
at x =

29

=
> y 2ax
> /

dy
O O
=
c


or
=
-jik --
1
+
= 496
y




Chapter 3 : Conic Sections

Ellipse = P ,
PF =
(x-de) +
y2 and PD = (4 - 1)
(2
&

Y e
a
+
p
= 1
PlacOSO ,
bsinO e =

(laesiy
form /
(standard form (
(parametric
One 1 for ellipses
,
c =
p p(a -




2x xY + =
( -


2aex + a
y
+




Y if foci ( 0)
a > b = 92 -29ex 2 ( 292x a y
+
= -
+ +

,
,




fe




↑
↓ ] <11-04




↑
b directrices ( = I a'll-e =
(11-24) +
ya
= 9


b

D' P(x y) ,
D
a


b " = a 2(1 2) -
" qxB +

y
=
1


· 1


bu




NOT
O
if b >a foci /0 Ibe) PF =
ePD PF' = DPD'
(C
,
,

A
! O
I
A
F F
-




1 -

42 , 0) (ae 0 ,
b
directrices y
=
I PD =
* -



/ DD =
G + C


A



a2= b2(1 22 -

PD +
PD =
2(8 -



x +
8 x) +




-
b PD +
PD' =
29




x =
- x =

ellipse : Ocel


1
(parametric form) of parabola e
Hyperbold eccentricity distance PF
:
e
=

= = ratio To PD

- Y =


Placosht ,
bsinht OR P(asect ,
btant /
hyperbola : I




I standard form / asymptotes

y = x
P
=

> 1 ,
for hyperbolas = ,
PF =
(x-de) +
y2 and PD (x-4)=




I See
*
foci (lae) e
y




(190 ,
0 e =




directrices : 1 =
I Ge e(l - +
) =
x-2aex + a +
y


b
=
a'(ec 1 -

e's'-2aex + a =
( -

29ex + a +
y



(C asymptotes :
y
= I B ( 2 1) - -




y =
a (e' -


1) = a'e' -

1




+page 1.
=



bu




Factorisation of Cubics : P= q =
(p 9) (p + +
pq
+
q'




Finding locus of midpoint :
Proving SQ =
eSP :




Point P lies on ellipse
-
Y = 1
, N is the foot of the perpendicular from P to the line ( =
8. Normal to
hyperbola -Yi = / at Plasect , btant) is allsint +
by =
(a' b)
+ .
Tant The normal meets (1-axis at Q.




M is the midpoint of PN .
Find locus of M .
Prove SQ = eSP ,
where S is the focus and e is the
eccentricity
.


Y x =
8 P(610S0 3 sinO) and N18 3 sinO Y x =
e b " = a (e- 1 e =

S




h
, ,


X




9. . .
3 M 8
/61030
+


Px 3sin0)
-




H
LN X
M :
,
Q :

y
=
0 ,
all sint =
(a
=
+
b) Tant SP =
ePN




- O ·
x
M)3COSO + 4 , 3sinO) x =
"e SP +e (a sect -


Q E of
-




3
-




let X 31098 + 4 Y 3 SinO SP =
de sect a
-




= =
, ,




coso =**, sino = so = -de
SP =
a lesect-1)


a "+ 9 'e' 1
Y
4)"
(X-
-




COSO sinO
+ = 1 =
-


af
1 SQ
=
eSP/
+ = : =
a cost




(X -

4) + Y =

9/
=
a -ae




SQ =
Ge/esect-1)