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# TOPIC 11: COORDINATE GEOMETRY II | MATHEMATICS FORM 6

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#### CONIC SECTIONS

Definition

Conic sections or conics are the sections whose ratios of the distance of the variables point from the fixed point to the distance, or the variable point from the fixed line is constant

#### TYPES OF CONIC SECTION

There are

i) Parabola

ii) Ellipse

iii) Hyperbola

IMPORTANT TERMS USED IN CONIC SECTION

I. FOCUS

This is the fixed point of the conic section.

For parabola

S → focus

For ellipse

S and S’ are the foci of an ellipse

II.  DIRECTRIX

This is the straight line whose distance from the focus is fixed.

For parabola

For ellipse

III.      ECCENTRICITY (e)

This is the amount ratio of the distance of the variable point from the focus to the distance of the variables point from the directrix.

For Parabola

For ellipse

IV.   AXIS OF THE CONIC

This is the straight line which cuts the conic or conic section symmetrically into two equal parts.

For parabola

X-axis is the point of the conic i.e. parabola

Also

Y-axis is the axis of the conic

FOR ELLIPSE

AB – is the axis (major axis) of the Conic i.e. (ellipse)

CD – is the axis (minor axis) the Conic i.e. (ellipse)

→An ellipse has Two axes is major and minor axes

V    FOCAL CHORD

This is the chord passing through the focus of the conic section.

For parabola

For ellipse

VI  LATUS RECTRUM

This is the focal cord which is perpendicular to the axis of the conic section.

For parabola

For Ellipse

Note:

Latus rectum is always parallel to the directrix

VII.  VERTEX

This is the turning point of the conic section.

For parabola

0 – is the vertex

For ellipse

Where V and V1 is the vertex of an ellipse

PARABOLA

This is the conic section whose eccentricity, e is one i.e. e = 1

For parabola

SP = MP

EQUATIONS OF THE PARABOLA

These are;

a) Standard equation

b) General equations

A. STANDARD EQUATION OF THE PARABOLA

1st case: Along the x – axis

·         Consider the parabola whose focus is S(a, 0) and directrix x = -a

Squaring both sides

Is the standard equation of the parabola

PROPERTIES

i) The parabola lies along x – axis

ii) Focus, s (a, o)

iii) Directrix x = -a

iv)  Vertex (0, 0) origin

Note:

PROPERTIES

1) The parabola lies along x – axis

2) Focus s (-a, o)

3) Directrix x = a

4) Vertex (o, o) origin

2nd case: along y – axis

Consider the parabola when focus is s (o, a) and directrix y = -a

·         Is the standard equation of the parabola along y – axis

PROPERTIES

i) The parabola lies along y – axis

ii) The focus s (o, a)

iii)  Directrix y = -a

iv)  Vertex (o, o) origin

Note;

Hence,    x2 = -4ay

PROPERTIES

i) The parabola lies along y – axis

ii) Focus s (o, -a)

iii) Directrix y = a

iv) Vertex (o, o)

GENERAL EQUATION OF THE PARABOLA

·         Consider the parabola whose focus is s (u, v) and directrix ax + by + c = 0

Is the general equation of the parabola

Where;

S (u, v) – is the focus

Examples:

1.  Find the focus and directrix of the parabola y2 = 8x

Solution

Given y2 = 8x

Comparing from

2.  Find the focus and the directrix of the parabola

y2 = -2x

Solution

Compare with

3.  Find the focus and directrix of x2 = 4y

Solution

4. Given the parabola x2 =

a) Find i) focus

ii) Directrix

iii) Vertex

b) Sketch the curve

Solution

i) Focus =

ii) Directrix, y = a

iii) Vertex

b) Curve sketching

5.  Find the equation of the parabola whose focus is (3, 0) and directrix

X = -3

Solution

Given focus (3, 0)

Directrix, x = -3

(3, 0) = (a, 0)

From

y2 = 4 (3)x

6.   Find the equation of the parabola whose directrix, y =

Solution

Given directrix, y =

Comparing with

7.  Find the equation of the parabola whose focus is (2, 0) and directrix, y = – 2.

Solution

Focus = (2, 0)

Directrix y = -2

8.   Find the equation of the parabola whose focus is (-1, 1) and directrix x = y

Solution

Given: focus = (-1, 1)

Directrix x = y

PARAMETRIC EQUATION OF THE PARABOLA
The parametric equation of the parabola  are given

X = at2 and y = 2at

Where;

t – is  a parameter

TANGENT TO THE PARABOLA

Tangent to the parabola, is the straight line which touches it at only one point.

Where, p – is the point of tangent or contact

CONDITIONS FOR TANGENT TO THE PARABOLA

a) Consider a line y = mx + c is the tangent to the parabola y2 = 4ax2. Hence the condition for tangency is obtained is as follows;

i.e.

b) Consider the line ax + by + c = is a tangent to the parabola y2 = 4ax Hence, the condition for tangency is obtained as follows;

i.e.

Examples

1.  Prove that the parametric equation of the parabola are given by

X = at2, and y = 2at

Solution

Consider the line

Y = mx + c is a tangent to the parabola y2 = 4ax. Hence the condition for tangency is given by y2 = 4ax

The parametric equation of the parabola of m is given as x = at2 and y = 2at

Where;

t – is a parameter

GRADIENT OF TANGENT OF THE PARABOLA

The gradient of tangent to the parabola can be expressed into;

i) Cartesian form

ii) Parametric form

i) IN CARTESIAN FORM

– Consider the tangent to the parabola y2 = 4ax Hence, from the theory.

Gradient of the curve at any = gradient of tangent to the curve at the point

ii) IN PARAMETRIC FORM

Consider the parametric equations of the parabola

i.e.

FORM SIX MATHEMATICS COORDINATE GEOMETRY II

EQUATION OF TANGENTS TO THE PARABOLA

These can be expressed into;

i) Cartesian form

ii) Parametric form

i) In Cartesian form

– Consider the tangent to the parabola y2 = 4ax at the point p (x, y)

Hence the equation of tangent is given by

ii) In parametric form

·         Consider the tangent to the parabola y2 = 4ax at the point p (at2, 2at)

Hence the equation of tangent is given by;

Examples

1. Show that the equation of tangent to the parabola y2 = 4ax at the point

2. Find the equation of tangent to the parabola y2 = 4ax at (at2, 2at)

NORMAL TO THE PARABOLA

Normal to the parabola is the line which is perpendicular at the point of tangency.

Where;

P is the point of tangency

GRADIENT OF THE NORMAL TO THE PARABOLA

This can be expressed into;·

i) Cartesian form

ii) Parametric form

i) In Cartesian form

– Consider the gradient of tangency in Cartesian form

i.e.

Let M = be gradient of the normal in Cartesian form but normal is perpendicular to tangent.

ii) In Parametric form

Consider the gradient of tangent in parametric form.

Let m be gradient of the normal in parametric form.

But

Normal is perpendicular to the tangent

EQUATION OF THE NORMAL TO THE PARABOLA

These  can  be expressed into;·
i) Cartesian form

ii) Parametric form

i) In Cartesian form

Consider the normal to the point y2= 4ax at the point p (x1, y1) hence the equation of the normal given by;

ii) In parametric form

Consider the normal to the parabola y2 = 4ax at the point p (at2, 2at). Hence the equation of the normal is given by;

Examples:

1.  Find the equation of the normal to the parabola y2 =  at the point

2. Show that the equation of the normal to the parabola y2 = 4ax at the point (at3, 2at) is

CHORD TO THE PARABOLA

·         This is the line joining two points on the parabola

Let m – be gradient of the chord
Hence

ii) GRADIENT OF THE CHORD IN PARAMETRIC FORM
Consider a chord to the parabola  at the points  and

EQUATION OF THE CHORD TO THE PARABOLA.

These can be expressed into;·
i) Cartesian form
ii) Parametric form

i) EQUATION OF THE CHORD IN PARAMETRIC FORM

– Consider the chord to the parabola y2 = 4ax at the points. Hence the equation of the chord is given by;

II. EQUATION OF THE CHORD IN CARTESIAN FORM.

Consider the chord to the parabola y2 = 4ax at the point P1(x1, y1) and P2 (x2, y2) hence the equation of the chord is given by

EXCERSICE.

1.  Show that equation of the chord to the parabola y2 = 4ax at (x1, y1) and (x2, y2) is

2.  Find the equation of the chord joining the points () and

3.  As the chord approaches the tangent at t1.deduce the equation of the tangent from the equation of the chord to the parabola y2 = 4ax.

THE LENGTH OF LATUS RECTUM

Consider the parabola

Now consider another diagram below

Therefore, the length of latus rectum is given by

EQUATION OF LATUS RECTUM
– The extremities of latus rectum are the points p1 (a, 2a) and

p2 (a1, -2a) as shown below

Therefore, the equation of latus rectum is given by

OPTICAL PROPERTY OF THE PARABOLA
Any ray parallel to the axis of the parabola is reflected through the focus. This property which is of considerable practical use in optics can be proved by showing that the normal line at the point ‘’p’’ on the parabola bisects the angle between  and the line  which is parallel to the axis of the parabola.
Angle of INCIDENCE and angle of REFLECTION are equal

– is the normal line at the point ‘p’ on the parabola
i.e.

Note that;  (QPS) Is an angle.

Examples
Prove that rays of height parallel to the axis of the parabolic mirror are reflected through the focus.

TRANSLATED PARABOLA

1.

– consider the parabola below

PROPERTIES.

I) The parabola is symmetrical about the line y = d through the focus
II) Focus,
III) Vertex,
IV) Directrix,

2.

– Consider the parabola below

PROPERTIES

I) the parabola is symmetrical about the line x = c, through the focus
II) Focus
III) Vertex,
IV) Directrix,

Examples

1. Show that the equation  represent the parabola and hence     find

i) Focus

ii) Vertex

iii) Directrix

iv) Length of latus rectum

Solution

Given;

FORM SIX MATHEMATICS COORDINATE GEOMETRY II

2. Shown that the equation x2 + 4x + 2 = y represents the parabola hence find its focus.

Solution

Given;

3.  Show that the equation x2 + 4x – 8y – 4 = 0 represents the parabola whose focus is at (-2, 1)

Solution

ELLIPSE

This is the conic section whose eccentricity e is less than one

I.e. |e| < 1

AXES OF AN ELLIPSE

An ellipse has two axes these are
i) Major axis
ii) Minor axis

1.  MAJOR AXIS

Is the one whose length is large

2.   MINOR AXIS

Is the one whose length is small

a)

b)

Where

AB – Major axis

PQ – Minor axis

EQUATION OF AN ELLIPSE
These are;
i) Standard equation

ii) General equation

1.   STANDARD EQUATION

– Consider an ellipse below;

1st CASE
Consider an ellipse along x – axis

FORM SIX MATHEMATICS COORDINATE GEOMETRY II

PROPERTIES

I) an ellipse lies along the x – axis (major axis)

ii) a > b

iii)

iv) Foci,

v)  Directrix

vi) Vertices, (a, o), (-a, o) along major axis

(0, b) (0, -b) along minor axis

vi) The length of the major axis l major = 2a

viii) Length of minor axis l minor = 2b

Note:

For an ellipse (a – b) the length along x – axis

B – is the length along y – axis

2nd CASE

·         Consider an ellipse along y – axis

PROPERTIES

i) An ellipse lies along y – axis (major axis)

ii) b > a

iii)

iv) Foci

v) Directrices

vi) Vertices:  = along major

= along minor as

vii) Length of the major axis L major = 2b

viii) Length of the minor axis   L minor = 2a

II. GENERAL EQUATION OF AN ELLIPSE

·         Consider an ellipse below y – axis

From

EXAMPLE
Given the equation of an ellipse
Find i) eccentricity
ii) Focus
iii) Directrices

Solution

Given

Compare from

Find the focus and directrix of an ellipse 9x+ 4y2 = 36

Solution

Given;

CENTRE OF AN ELLIPSE
This is the point of intersection between major and minor axes

·         O – Is the centre of an ellipse

A – Is the centre of an ellipse

DIAMETER OF AN ELLIPSE.

This is any chord passing through the centre of an ellipse

Hence  – diameter (major)
– Diameter (minor)
Note:
i) The equation of an ellipse is in the form of

ii) The equation of the parabola is in the form of

iii) The equation of the circle is in the form of

PARAMETRIC EQUATIONS OF AN ELLIPSE

The parametric equations of an ellipse are given as

And

Where

θ – Is an eccentric angle

Recall

TANGENT TO AN ELLIPSE
This is the straight line which touches the ellipse at only one point

Where;
P – Is the point of tangent or contact
Condition for tangency to an ellipse
Consider the line b = mx + c is the tangent to an ellipse

Examples
Show that, for a line  to touch the ellipse        Then,

GRADIENT OF TANGENT TO AN ELLIPSE
This can be expressed into;
i) Cartesian form
ii) Parametric form

1. GRADIENT OF TANGENT IN CARTESIAN FORM

– Consider an ellipse

Differentiate both sides with w.r.t x

ii. GRADIENT OF TANGENT IN PARAMETRIC FORM

– Consider the parametric equation of an ellipse

EQUATION OF TANGENT TO AN ELLIPSE

These can be expressed into;
i) Cartesian form
ii) Parametric form

I.   Equation of tangent in Cartesian form

– Consider the tangent an ellipse

Hence, the equation of tangent is given by

EQUATION OF TANGENT IN PARAMETRIC FORM.

Consider the tangent to an ellipse  At the point

Hence the equation of tangent is given by

Note
1.

2.

EXERCISE
i. Show that the equation of tangent to an ellipse
ii.  Show that the equation of tangent to an ellipse
iii.  Show that the gradient of tangent to an ellipse
NORMAL TO AN ELLIPSE
Normal to an ellipse perpendicular to the tangent at the point of tangency.

Where: p is the point of tangency

GRADIENT OF THE NORMAL TO AN ELLIPSE.

This can be expressed into two
i) Cartesian form
ii) Parametric

I) IN CARTESIAN FORM

– Consider the gradient of the tangent in Cartesian form

But normal tangent

II) IN PARAMETRIC FORM
Consider the gradient of tangent in parametric form

Let m = slope of the normal in parametric form

EQUATION OF THE NORMAL TO AN ELLIPSE
This can be expressed into;
(i) Cartesian form
(ii)  Parametric form

I. IN CARTESIAN FORM

– consider the normal to an ellipse

Hence the equation of the normal is given by

FORM SIX MATHEMATICS COORDINATE GEOMETRY II

II) IN PARAMETRIC FORM
Consider the normal to an ellipse

Hence the equation of the normal is given by

Examples
·         Show that the equation of the normal to an ellipse

CHORD OF AN ELLIPSE.

This is the line joining any two points on the curve ie (ellipse)

GRADIENT OF THE CHORD TO AN ELLIPSE.

This can be expressed into
i) Cartesian form
ii) Parametric form

I.  IN CARTESIAN FORM

– Consider the point A (x1, y1) and B (x2, y2) on the ellipse  hence the gradient of the cord is given by

II. IN PARAMETRIC FORM

Consider the points A  and B  on the ellipse  Hence the gradient of the chord is given by;

EQUATION OF THE CHORD TO AN ELLIPSE

These can be expressed into
i) Cartesian form
ii) Parametric form
I:  IN CARTESIAN FORM.

Consider the chord the ellipse at the point A (x1, y1) and B(x2,y2). Hence the equation of the chord is given by;

II.   IN PARAMETRIC FORM.

Consider the chord to an ellipse  at the points. Hence the equation of the chord is given by

FOCAL CHORD OF AN ELLIPSE.

This is the chord passing through the focus of an ellipse

Where  = is the focal chord
Consider the points A and B are respectively  Hence
Gradient of AS = gradient of BS
Where s = (ae, o)

DISTANCE BETWEEN TWO FOCI.

Consider the ellipse below;

2

2

2

Where a = is the semi major axis

e = is the eccentricity

DISTANCE BETWEEN DIRECTRICES.

Consider the ellipse below

=

=

Where a – is the semi major axis
e – is the eccentricity

LENGTH OF LATUS RECTUM.

Consider the ellipse below

IMPORTANT RELATION OF AN ELLIPSE

Consider an ellipse  below

ECCENTRIC ANGLE OF ELLIPSE

.This is the angle introduced in the parametric equation of an ellipse
I.e
Where      – is an eccentric angle
CIRCLES OF AN ELLIPSE
These are 1) Director Circle
2) Auxiliary circle

1.   DIRECTOR CIRCLE
– This is the locus of the points of intersection of the perpendicular tangents.

Consider the line  is tangent to the ellipse
Hence

2.  AUXILIARY CIRCLE
– This is the circle whose radius is equal to semi – major axis

Using Pythagoras theorem

a – is the radius of the auxiliary circle

FORM SIX MATHEMATICS COORDINATE GEOMETRY II

CONCENTRIC ELLIPSE.

These are ellipse whose centre are the same.

The equations of centric ellipse are;

Where  a and b semi – major and semi – minor axes of the small ellipse

A and B are the semi – major and semi – minor axes of the large ellipse

A – a = B – b

A – B = a – b·
Is the condition for concentric ellipse

TRANSLATED ELLIPSE

This is given by the equation

A. PROPERTIES

i) An ellipse lies along x – axis
ii) a > b
iii) Centre (h, k)
iv) Vertices
v) Eccentricity,
vi) Foci
vii) Directrices
B. PROPERTIES
i) An ellipse has along y – axis
ii) b > a
iii) Centre (h, k)
iv) Vertices
v) Eccentricity
vi) Foci

Examples
Show that the equation 4x2 – 16x + 9y2 + 18y – 11 = 0 represents an    ellipse and hence find i) centre ii) vertices iii) eccentricity iv) foci v) directrices.

Solution

FORM SIX MATHEMATICS COORDINATE GEOMETRY II

III.      HYPERBOLA

This is the conic section whose eccentricity ‘’e’’ is greater than one       ( e > 1)

The hyperbola has two foci and two directrices

Where S and S’ are the foci of the hyperbola hence

Where e > 1

EQUATION OF THE HYPERBOLA

There are;
i) Standard equation
ii) General equation

1.    STANDARD EQUATION OF THE HYPERBOLA

Consider