Conic SectionsImagine you slice through a cone at different angles

circle

ellipse

parabola

rectangular hyperbola

You could get a cross-section which is a:

These shapes are all important functions in Mathematics and occur in fields as diverse as the motion of planets to the optimum design of a satellite dish.

In FP1 you consider the algebra & geometry of 2 of these – the parabola and rectangular hyperbola

Parabola.aggThe Parabola

Q is the point horizontally in line horizontally with P on the line x = -a

The restriction that P can move such that QP = PS is the focus-directrix property

The locus of points for P is a parabolaPQ

S(a,0)

x = -a

P can move such that QP=PS …

Consider a point P that can move according to a rule:

The point S has coordinates (a,0)

The point S(a,0) is called the focus

The line x = -a is called the directrix

The Cartesian equation is y2 = 4ax

Sub in

Figure 1

WB14 Figure 1 shows a sketch of the parabola C with equation (a) The point S is the focus of C. Find the coordinates of S.

(b) Write down the equation of the directrix of C.

Figure 1 shows the point P which lies on C, where y > 0, and the point Q which lies on the directrix of C. The line segment QP is parallel to the x-axis.(c) Given that the distance PS is 25, write down the distance QP,

(d) find the coordinates of P,

(e) find the area of the trapezium OSPQ.

xy 362

axy 42 where the focus is S(a,0)and the directrix has equation x = -a

Coordinates of S are (9,0)

Equation of directrix x = -9

Focus-directrix property: PS = PQ

QP = 25

259 16

16x xy 362 24 yCoordinates of P are (16,24)

24

242259Area

9

408

WB15 Figure 1 shows a sketch of part of the parabola with equation y2 = 12x .The point P on the parabola has x-coordinate 3

1

The points A and B lie on the directrix of the parabola.The point A is on the x-axis and the y-coordinate of B is positive.Given that ABPS is a trapezium,(b) calculate the perimeter of ABPS.

Figure 1The point S is the focus of the parabola.(a) Write down the coordinates of S.

axy 42 where the focus is S(a,0)

Directrix has equation x = -a

Coordinates of S are (3,0)3

Sub in 31x xy 122 2 y

3

313

2

313

Focus-directrix property

313 PBPS

2 AB3 OSAO

Perimeter = 3214

at P

Eg a curve has parametric equations ,

Parametric functions

Some simple-looking curves are hard to describe with a Cartesian equation.Parametric equations, where the values of x and y are determined by a 3rd

variable t, can be used to produce some intricate curves with simple equations.

tx 2 2ty

t -3 -2 -1 0 1 2 3

x

y

-6

Complete the table and sketch the curve

9

-4

4

-2

1

0

0

2

1

4

4

6

9

NB: you can still find the Cartesian equation of a function defined parametrically…

tx 22xt

Sub in 2ty 2

2

xy 2

41 x

Problem solving with parametric functionsEg a curve has parametric equations ,1tx 24 ty The curve meets the x-axis at A and B, find their coordinates

A B

At A and B, 0y 42 t

2 t

13, x

Coordinates are (-3,0) and (1,0)

Eg a curve has parametric equations ,2tx ty 4The line meets the curve at A. Find the coordinates of A04 yx

0442 tt

02 2 t2 t

84 ,A

Substitute the expressions for x and y in terms of t to solve the equations simultaneously

Solve

Substitute value of t back into expressions for x and y

Find values of t at A and B

Substitute values of t back into expression for x

The parametric form of a parabola is , 2atx aty 2 Does this fit with its Cartesian equation?

Sub into axy 42 22 42 ataat 2222 44 tata which is true!

Exam questions sometimes involve the parabola’s parametric form…

WB16 The parabola C has equation y2 = 20x.(a) Verify that the point P(5t2 ,10t) is a general point on C.

The point A on C has parameter t = 4.The line l passes through A and also passes through the focus of C.(b) Find the gradient of l.

Sub in tt 105 2 , xy 202 22 52010 tt

4t 4080,A

axy 42 has focus S(a,0)

05,S5 axy 202

05,S

4080,A

75

40

7540 of Gradient l 15

8l

22 100100 tt

The equation of the straight line with gradient m that passes through is

(b) Show that the equation of the tangent to C at P(12t2, 24t) is x − ty + 12t2 = 0.

The tangent to C at the point (3, 12) meets the directrix of C at the point X.(c) Find the coordinates of X

WB17 The parabola C has equation y2 = 48x.The point P(12t2, 24t) is a general point on C.(a) Find the equation of the directrix of C.

axy 42 where the focus is S(a,0)and the directrix has equation x = -a

Equation of directrix x = -12

Sub 212tx t1 at P212

32tdx

dy

Giving tangent

)( 11 xxmyy ),( 11 yx

)( 21 1224 txty t 22 1224 txtty

012 2 ttyx

123,

X

Comparing (3,12) with (12t2, 24t) 21 t at (3,12)

Sub n equation of tangent 21t 032

1 yxWhen this intersects directrix x = -12

0312 21 y 18 y

Coordinates of X are (-12,-18)

12xDirectrix

xy 482 21

34 xy 21

32 xdxdy

x32

The Rectangular Hyperbola

The rectangular hyperbola also has a focus-directrix property, but it is beyond the scope of FP1. You only need to know that:

The Cartesian equation is xy = c2

The parametric form of a parabola is , ctx tcy

Problems involving rectangular hyperbola usually require to find the equation of the tangent or normal

for functions given explicitly or in terms of c

Sub ctx 2

1t

22

2

tcc

dxdy

2cxy 12 xcy 22 xcdxdy

2

2

xc

WB19 The rectangular hyperbola H has equation xy = c2, where c is a positive constant. The point A on H has x-coordinate 3c.(a) Write down the y-coordinate of A.

(b) Show that an equation of the normal to H at A is

2cxy with general point tcct,

3 t 3 coordinate cy

Sub91

at A 22

3cc

dxdy

33 cc,

9 normal of Gradient The equation of the straight line with gradient m that passes through is

Giving normal

)( 11 xxmyy ),( 11 yx

)( cxy c 393

cxcy 81273 cxy 80273

cxy 80273

(c) The normal to H at A meets H again at the point B. Find, in terms of c, the coordinates of B.

2cxy xcy 2

Sub in cxy 80273 cxx

c 802723

Solve and simultaneously to find points of intersection

cxxc 80273 22

038027 22 ccxx

0 3 cx x27 c

27cx at B

Given solution x = 3c

cy 27

Coordinates of B are cc 27 27 ,

2cxy cxy 80273

using xcy 2

2cxy 12 xcy 22 xcdxdy

2

2

xc

WB20 The point P , t ≠ 0, lies on the tt 66 ,

(a) Show that an equation for the tangent to H at P is

(b) The tangent to H at the point A and the tangent to H at the point B meet at the point (−9, 12). Find the coordinates of A and B.

rectangular hyperbola H with equation xy = 36.

ttxy 121

2

The equation of the straight line with gradient m that passes through is

Sub 2

1t

at Ptdx

dy t

6

6

Giving tangent

)( 11 xxmyy ),( 11 yx

)( txytt 6216

ttt xy 6162

tt 66 ,

ttxy 121

2

Sub in 129, ttxy 121

2

tt129

212

tt 12912 2

0344 2 tt03624 2 ttt

0123122 ttt 01232 tt

21

23 , t

Sub in 63 49 ,,, tt 66 ,

2cxy 12 xcy 22 xcdxdy

2

2

xc

WB18 The rectangular hyperbola H has equation xy = c2, where c is a constant.

tcct,

(a) Show that the tangent to H at P has equation t2y + x = 2ct.

The point P is a general point on H.

The tangents to H at the points A and B meet at the point (15c, –c).(b) Find, in terms of c, the coordinates of A and B.

The equation of the straight line with gradient m that passes through is

2cxy 12 xcy 22 xcdxdy

Sub 2

1t

at P 22

ctc

dxdy

Giving tangent

)( 11 xxmyy ),( 11 yx

)( ctxytt

c 21

ctxctyt 2

tcct,

ctxyt 22

Sub in cc ,15 ctxyt 22

ctcct 2152 01522 cctct

053 tt5 3 ,t

Sub values in tcct,

53 5 3 cc c,-- c ,,

2

2

xc

Parabola Rectangular hyperbola

Standard form

Parametric form

Foci

Directrices

axy 42 2cxy

atat 2 2 , tcct ,

0 ,a

ax

Not required

Not required

Formulae sheet facts

Sub ctx 2

1t

22

2

tcc

dxdy

2cxy 12 xcy 22 xcdxdy

2

2

xc

Sub 2atx t1

ta

adxdy

axy 42 21

2 xay 21 xa

dxdy

xa

Obtaining the gradient as a function of tParabola

Rectangular hyperbola