Transcript B - Runnymede Mathematics Department

Conic Sections
Imagine you slice through a cone at different angles
You could get a cross-section which is a:
circle
ellipse
parabola
rectangular hyperbola
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
The Parabola
Parabola.agg
Consider a point P that can move according to a rule:
Q is the point horizontally in line horizontally with P on the line x = -a
The point S has coordinates (a,0)
P
Q
P can move such that QP=PS …
The locus of points for P is a parabola
The point S(a,0) is called the focus
The line x = -a is called the directrix
S(a,0)
The restriction that P can move such that
QP = PS is the focus-directrix property
The Cartesian equation is y2 = 4ax
x = -a
WB14 Figure 1 shows a sketch of the parabola C with equation y 2  36 x
(a) The point S is the focus of C. Find the coordinates of S.
y 2  4ax where the focus is S(a,0)
Figure 1
and the directrix has equation x = -a
25
16
9
Coordinates of S are (9,0)
24
(b) Write down the equation of the directrix of C.
Equation of directrix x = -9
9
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,
Focus-directrix property: PS = PQ
QP = 25
(d) find the coordinates of P,
Sub x  16 in y 2  36 x  y  24
Coordinates of P are (16,24)
(e) find the area of the trapezium OSPQ.
Area 
9  25
 24  408
2
WB15 Figure 1 shows a sketch of part of the parabola with equation y2 = 12x .
1
The point P on the parabola has x-coordinate 3
Figure 1
The point S is the focus of the parabola.
(a) Write down the coordinates of S.
y 2  4ax where the focus is S(a,0)
3 31
3 31
2
Coordinates of S are (3,0)
3
3
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.
Directrix has equation x = -a
 AO  OS  3
Sub x 
1
3
in y 2  12 x  y  2 at P
 AB  2
Focus-directrix property
 PS  PB  3
1
3
Perimeter = 14 32
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.
Eg a curve has parametric equations x  2t , y  t
Complete the table and sketch the curve
t
-3
-2
-1
0
1
2
3
x
-6
-4
-2
0
2
4
6
y
9
4
1
0
1
4
9
NB: you can still find the Cartesian equation
of a function defined parametrically…
x  2t  t 
x
2
2
 x 1 2
2
Sub in y  t  y     x
2 4
2
Problem solving with parametric functions
Eg a curve has parametric equations x  t  1 , y  4  t 2
The curve meets the x-axis at A and B, find their coordinates
2
At A and B, y  0  t  4
Find values of t at A and B
Substitute values of t
back into expression for x
A
B
 t  2
 x  3,1
Coordinates are (-3,0) and (1,0)
Eg a curve has parametric equations x  t , y  4t
2
The line x  y  4  0 meets the curve at A. Find the coordinates of A
 t 2  4t  4  0
 t  2  0
Substitute the expressions for x and y in terms
of t to solve the equations simultaneously
2
 t  2
 A4,8
Solve
Substitute value of t back
into expressions for x and y
The parametric form of a parabola is x  at , y  2at
Does this fit with its
Cartesian equation?
2
   4a t
2
Sub into y  4ax  2at   4a at
2
2 2
2
 4a 2t 2 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.


   100t
2
Sub 5t ,10t in y  20 x  10t   20 5t
2
2
2
2
 100t 2
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.
A80,40
40
t  4  A80,40
S 5,0
75
y  4ax has focus S(a,0)
2
y 2  20 x  a  5  S 5,0
40
Gradient of l  75
 158
l
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.
y 2  4ax where the focus is S(a,0)
and the directrix has equation x = -a
Equation of directrix x = -12
(b) Show that the equation of the tangent to C at P(12t2, 24t) is x − ty + 12t2 = 0.
dy
 21
y  48 x  y  4 3 x   2 3 x  2
dx
dy
2
 2 123t 2  1t at P
Sub x  12t 
dx
1
2
2
The equation of the straight line with
gradient m that passes through ( x1 , y1 )
is y  y1  m( x  x1 )
3
x
Giving tangent
The tangent to C at the point (3, 12)
meets the directrix of C at the point X.
(c) Find the coordinates of X
y  24t  1t ( x  12t 2 )
 ty  24t 2  x  12t 2
 x  ty  12t 2  0
Directrix
x  12
3,12
Comparing (3,12) with (12t2, 24t) t 
at (3,12)
 x  21 y  3  0
When this intersects directrix x = -12
Sub t 
X
1
2
1
2
n equation of tangent
 12  21 y  3  0  y  18
Coordinates of X are (-12,-18)
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 x  ct , y 
c
t
Problems involving rectangular hyperbola usually
require to find the equation of the tangent or normal
for functions given explicitly or in terms of c
dy
c2
2 2
xy  c  y  c x   c x   2
dx 2
x
dy
c
1
Sub x  ct 
 2 2  2
dx
ct
t
2
2 1
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.
(c) The normal to H at A meets H
xy  c 2 with general point ct, ct 
 t  3  y  coordinate
c
3
(b) Show that an equation of the
normal to H at A is 3 y  27 x  80c
again at the point B. Find, in
terms of c, the coordinates of B.
Solve xy  c 2 and 3 y  27 x  80c
simultaneously to find points of intersection
2
dy
c
2 2
 c x   2
xy  c  y  c x 
dx
x
2
dy
c
1

Sub 3c, 3c  
at A


2
dx
3c 
9
2 1
2
 Gradient of normal  9
The equation of the straight line with
gradient m that passes through ( x1 , y1 )
is y  y1  m( x  x1 )
Giving normal
y  3c  9( x  3c)
 3 y  c  27 x  81c
 3 y  27 x  80c
xy  c  y 
2
Sub in
c2
x
3 y  27 x  80c
 3cx  27 x  80c
2
 3c 2  27 x 2 80cx
 27 x 2 80cx  3c 2  0
Given solution x = 3c
 x  3c 27 x  c  0
 x   27c at B
 y  27c using y  cx
2
Coordinates of B are
 27c ,  27c 


WB20 The point P 6t , 6t , t ≠ 0, lies on the
rectangular hyperbola H with equation xy = 36.
(a) Show that an equation for the tangent to H at P is y  
xy  c 2  y  c 2 x 1

Sub 6t ,
6
t

2
dy
c
2 2

 c x   2
dx
x
6
1
dy
t
at P




2
t
dx
6t
The equation of the straight line with
gradient m that passes through ( x1 , y1 )
is y  y1  m( x  x1 )
Giving tangent
y  6t   t12 ( x  6t )
 y  6t   t12 x  6t
 y   t1 x  12t
2
1
t2
x  12t
(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.
Sub  9, 12 in y   t12 x  12t
 12  t92  12t
 12t 2  9  12t
 4t 2  4t  3  0
 4t 2  2t  6t  3  0
 2t 2t  1  32t  1  0
 2t  32t  1  0
 t  32 , 21

Sub in 6t ,
6
t
  9,4 , 3,6
WB18 The rectangular hyperbola H has equation xy = c2, where c is a constant.

The point P ct,
c
t
 is a general point on H.
(a) Show that the tangent to H at P has equation t2y + x = 2ct.
xy  c  y  c x
2

Sub ct,
2
c
t

2
dy
c
2 2

 c x   2
dx
x
1
dy
c2
1




at P
dx
t2
ct 2
The equation of the straight line with
gradient m that passes through ( x1 , y1 )
is y  y1  m( x  x1 )
Giving tangent
y  ct   t12 ( x  ct )
 t 2 y  ct   x  ct
 t y  x  2ct
2
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.
Sub 15c,
 c in t 2 y  x  2ct
 ct 2  15c  2ct
 ct 2  2ct  15c  0
 t  3t  5  0
 t  3,  5


 3c, 3c , -5c,- 5c 
Sub values in ct,
c
t
Formulae sheet facts
Standard form
Parametric form
Parabola
Rectangular
hyperbola
y 2  4ax
xy  c 2
at
ct , ct 
2
, 2at

Foci
a , 0
Not required
Directrices
x  a
Not required
Obtaining the gradient as a function of t
Parabola
a
dy
 21
y  4ax  y  2 a x   a x 
dx
x
1
2
2
2
Sub x  at 
dy
a 1


dx
at t
Rectangular hyperbola
2 1
xy  c  y  c x
2
c2
dy
2 2
  c x   2
x
dx
dy
c2
1
Sub x  ct 
 2 2  2
dx
ct
t