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MA1200 Basic Calculus and Linear Algebra I
Lecture Note 1

Coordinate Geometry and Conic Sections

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MA1200 Basic Calculus and Linear Algebra I
Lecture Note 1: Coordinate Geometry and Conic Sections

Topic Covered

• Two representations of coordinate systems: Cartesian coordinates [ሺ‫ݕ ,ݔ‬ሻcoordinates] and Polar coordinates [ሺ‫ߠ ,ݎ‬ሻ-coordinates].

• Conic Sections: Circle, Ellipse, Parabola and Hyperbola.
• Classify the conic section in 2-D plane
General equation of conic section
Identify the conic section in 2-D plane
- Useful technique: Rotation of Axes
- General results

φ

MA1200 Basic Calculus and Linear Algebra I
Lecture Note 1: Coordinate Geometry and Conic Sections

Representations of coordinate systems in 2-D
There are two different types of coordinate systems used in locating the position of a point in 2-D.
First representation: Cartesian coordinates
We describe the position of a given point by considering the (directed) distance between the point and ‫-ݔ‬axis and the distance between the point and ‫-ݕ‬axis.
‫ݕ‬

0

ܽ

ܲ ൌ ሺܽ, ܾሻ
ܾ

‫ݔ‬

Here, ܽ is called “‫-ݔ‬coordinate” of ܲ and ܾ is called “‫-ݕ‬coordinate” of ܲ.

χ

MA1200 Basic Calculus and Linear Algebra I
Lecture Note 1: Coordinate Geometry and Conic Sections

ܲଶ ൌ ሺ‫ݔ‬ଶ , ‫ݕ‬ଶ ሻ
ܲଵ ൌ ሺ‫ݔ‬ଵ , ‫ݕ‬ଵ ሻ

Given two points ܲଵ ൌ ሺ‫ݔ‬ଵ , ‫ݕ‬ଵ ሻ and ܲଶ ൌ ሺ‫ݔ‬ଶ , ‫ݕ‬ଶ ሻ, we learned that

• the distance between ܲଵ and ܲଶ : ܲଵ ܲଶ ൌ ඥሺ‫ݔ‬ଶ െ ‫ݔ‬ଵ ሻଶ ൅ ሺ‫ݕ‬ଶ െ ‫ݕ‬ଵ ሻଶ ,
• the slope of line segment joining ܲଵ and ܲଶ :
‫ݕ‬ଶ െ ‫ݕ‬ଵ
,
݈ܵ‫ ݁݌݋‬ൌ ݉ ൌ
‫ݔ‬ଶ െ ‫ݔ‬ଵ
• the equation of the straight line ܲଵ ܲଶ :
‫ݕ‬ଶ െ ‫ݕ‬ଵ
‫ ݕ‬െ ‫ݕ‬ଵ
ൌ݉ൌ
ሺܲ‫ ݐ݊݅݋‬െ ‫݉ݎ݋݂ ݁݌݋݈ݏ‬ሻ,
‫ ݔ‬െ ‫ݔ‬ଵ
‫ݔ‬ଶ െ ‫ݔ‬ଵ
• The equation can be expressed in the following form:
‫ ݕ‬ൌ ݉‫ ݔ‬൅ ܿ ሺ‫ ݁݌݋݈ݏ‬െ ݅݊‫݉ݎ݋݂ ݐ݌݁ܿݎ݁ݐ‬ሻ where ݉ and ܿ are the slope and y-intercept respectively.

ψ

MA1200 Basic Calculus and Linear Algebra I
Lecture Note 1: Coordinate Geometry and Conic Sections

Example 1

Find the equation of straight line ‫ ܮ‬which satisfies each of the following conditions: (a) Passing through ܲ ൌ ሺ3,4ሻ and ܳ ൌ ሺ1, െ1ሻ
(b) Perpendicular to the line ‫ܮ‬ଵ : 3‫ ݔ‬െ 2‫ ݕ‬൅ 5 ൌ 0 and cuts the ‫-ݔ‬axis at
ሺ5,0ሻ.

☺Solution:

(a) The slope of ܲܳ ൌ

ିଵିସ
ଵିଷ

ൌ . The equation of ܲܳ is then given by



‫ݕ‬െ4 5
ൌ ⇒ 2ሺ‫ ݕ‬െ 4ሻ ൌ 5ሺ‫ ݔ‬െ 3ሻ ⇒ 5‫ ݔ‬െ 2‫ ݕ‬െ 7 ൌ 0.
‫ݔ‬െ3 2

IDEA: In order to find the equation of ‫ ,ܮ‬we have to find the slope of ‫.ܮ‬
Since ‫ ܮ‬and ‫ܮ‬ଵ are perpendicular, then we can obtain this by firsting finding the slope of ‫ܮ‬ଵ .

(b)

ω

MA1200 Basic Calculus and Linear Algebra I
Lecture Note 1: Coordinate Geometry and Conic Sections

To obtain the slope of ‫ܮ‬ଵ , one has to express the equation of ‫ܮ‬ଵ in the slope-intercept form:
3
5
3‫ ݔ‬െ 2‫ ݕ‬൅ 5 ൌ 0 ⇒ 2‫ ݕ‬ൌ 3‫ ݔ‬൅ 5 ⇒ ‫ ݕ‬ൌ ‫ ݔ‬൅ .
2
2

We get the slope of ‫ܮ‬ଵ is ݉ଵ ൌ .


Since the line ‫ ܮ‬and ‫ܮ‬ଵ are perpendicular, we have
3
2
݉ ൈ ݉ଵ ൌ െ1 ⇒ ݉ ൌ െ1 ⇒ ݉ ൌ െ .
2
3
(Here, ݉ is the slope of ‫)ܮ‬

Then the equation of ‫ ܮ‬is found to be
‫ݕ‬െ0
2
ൌ െ ⇒ 3‫ ݕ‬ൌ െ2ሺ‫ ݔ‬െ 5ሻ ⇒ 2‫ ݔ‬൅ 3‫ ݕ‬െ 10 ൌ 0.
‫ݔ‬െ5
3

ϊ

MA1200 Basic Calculus and Linear Algebra I
Lecture Note 1: Coordinate Geometry and Conic Sections

Second coordinates: Polar coordinates
Alternatively, one can describe the position of a point by considering the distance between the point ܲ and the origin (‫ )ݎ‬and the (directed) angle between ܱܲ and the positive ‫-ݔ‬axis ሺߠሻ.
‫ݕ‬

ܾ
0

Note:

ߠ ൐ 0: anti-clockwise direction
ߠ ൏ 0: clockwise direction

‫ݎ‬

ߠ

ܲ ൌ ሺ‫ߠ ,ݎ‬ሻ or ሺܽ, ܾሻ
ܽ

‫ݔ‬

The relations between Polar ሺ‫ߠ ,ݎ‬ሻ and Cartesian ሺ‫ݕ ,ݔ‬ሻ Coordinates are given by ‫ݕ‬
ඥ‫ ݔ‬ଶ ൅ ‫ ݕ‬ଶ , tan ߠ ൌ .
‫ ݔ‬ൌ ‫ ݎ‬cos ߠ ,
‫ ݕ‬ൌ ‫ ݎ‬sin ߠ , ‫ ݎ‬ൌ
‫ݔ‬

ϋ

MA1200 Basic Calculus and Linear Algebra I
Lecture Note 1: Coordinate Geometry and Conic Sections

Example 2

Given the points ‫ ܣ‬ൌ ሺ1,45°ሻ, ‫ ܤ‬ൌ ሺ2,150°ሻ, ‫ ܥ‬ൌ ሺ3, െ60°ሻ and ‫ ܦ‬ൌ ሺ2,440°ሻ. Locate the points in the ‫-ݕݔ‬plane. Find the corresponding Cartesian coordinates of these points.
☺Solution:

‫ ܤ‬ൌ ሺ2,150°ሻ

‫ݕ‬

0

‫ ܦ‬ൌ ሺ2,440°ሻ

‫ ܣ‬ൌ ሺ1,45°ሻ

‫ ܥ‬ൌ ሺ3, െ60°ሻ

‫ݔ‬

The corresponding Cartesian coordinates of these points are given by
1 1
‫ ܣ‬ൌ ሺ1 cos 45° , 1 sin 45°ሻ ൌ ൬ , ൰ ,
‫ ܤ‬ൌ ሺ2 cos 150° , 2 sin 150°ሻ ൌ ൫െ√3, 1൯,
√2 √2
3 3√3
ቇ,
‫ ܦ‬ൌ ሺ2 cos 440° , 2 sin 440°ሻ ൎ ሺ0.35,1.97ሻ.
‫ ܥ‬ൌ ሺ3 cos െ60° , 3 sin െ60°ሻ ൌ ቆ , െ
2
2

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MA1200 Basic Calculus and Linear Algebra I
Lecture Note 1: Coordinate Geometry and Conic Sections

Example 3

Let ‫ ܣ‬ൌ ሺ√3, 3ሻ, ‫ ܤ‬ൌ ሺ1, െ1ሻ, ‫ ܥ‬ൌ ሺെ2, െ2ሻ. Find the corresponding polar coordinates of ‫( .ܥ ,ܤ ,ܣ‬Here, choose െ180° ൏ ߠ ൑ 180°)
☺Solution:
Tips: It is always a good idea to locate the points on ‫ ݕ ,ݔ‬so that you can obtain the values of ‫ ݎ‬and ߠ correctly.
‫ݕ‬

‫ ܥ‬ൌ ሺെ2, െ2ሻ

ύ

0

‫ ܣ‬ൌ ൫√3, 3൯

‫ ܤ‬ൌ ሺ1, െ1ሻ

‫ݔ‬

MA1200 Basic Calculus and Linear Algebra I
Lecture Note 1: Coordinate Geometry and Conic Sections

For Point ‫ܣ‬

‫ݎ‬஺ ൌ ට൫√3൯ ൅ 3ଶ ൌ √12. tan ߠ஺ ൌ



√ଷ

For Point ‫ܤ‬



‫ݎ‬஺

ൌ √3 ⇒ ߠ஺ ൌ 60°.

‫ݎ‬஻ ൌ ඥሺെ1ሻଶ ൅ 1ଶ ൌ √2. tan ߠ஻ ൌ

ିଵ


For Point ‫ܥ‬

ൌ െ1 ⇒ ߠ஻ ൌ െ45°.

‫ݎ‬௖ ൌ ඥ2ଶ ൅ ሺെ2ሻଶ ൌ √8.

tan ߙ஼ ൌ ൌ 1 ⇒ ߙ஼ ൌ 45°.



So ߠ஼ ൌ െሺ180° െ 45°ሻ ൌ െ135°.

∴ ‫ ܣ‬ൌ ൫√12, 60°൯,

υτ

‫ݎ‬஻

‫ ܥ‬ൌ ሺെ2, െ2ሻ

‫ ܤ‬ൌ ൫√2, െ45°൯,

‫ ܣ‬ൌ ൫√3, 3൯
ߠ஺

ߠ஻

‫ ܤ‬ൌ ሺ1, െ1ሻ

ߙ஼

‫ݎ‬஼

ߠ஼

‫ ܥ‬ൌ ൫√8, െ135°൯.

MA1200 Basic Calculus and Linear Algebra I
Lecture Note 1: Coordinate Geometry and Conic Sections

☺Note: In the remaining section, we will adopt the Cartesian coordinates system unless otherwise specified.
Conic Sections
Conic sections, by definition, are curves that result from intersecting a right circular cone with a plane. The following figure shows the four conic sections:

Circle

υυ

Ellipse

Parabola

MA1200 Basic Calculus and Linear Algebra I
Lecture Note 1: Coordinate Geometry and Conic Sections

Hyperbola

Conic Section #1: Circle

Given a fixed point ‫ ,ܥ‬a circle is a set of all points such that the distance between the point and ‫ ܥ‬is always fixed.

(*‫ ܥ‬is called centre of the circle, the fixed distance is called radius of the circle and is denoted by ‫).ݎ‬
ܲ’

υφ

‫ݎ‬

‫ݎ‬

ܲ

‫ܥ‬

MA1200 Basic Calculus and Linear Algebra I
Lecture Note 1: Coordinate Geometry and Conic Sections

The equation of circle

The circle with centre at the origin ܱ ൌ ሺ0,0ሻ
For any point ܲ lying on the circle, we have

ܱܲ ൌ ‫ ⇒ ݎ‬ඥሺ‫ ݔ‬െ 0ሻଶ ൅ ሺ‫ ݕ‬െ 0ሻଶ ൌ ‫ݎ‬
⇒‫ ݔ‬൅‫ ݕ‬ൌ‫ݎ‬






which is the equation of circle (with centre origin).

‫ݕ‬

ܱ

ܲ ൌ ሺ‫ݕ ,ݔ‬ሻ

‫ݎ‬

Using similar method, we can derive the following general case:
Equation of Circle (General Case)

The equation of circle with centre ‫ ܥ‬ൌ ሺ݄, ݇ሻ and radius ‫ ݎ‬is given by

ሺ‫ ݔ‬െ ݄ሻଶ ൅ ሺ‫ ݕ‬െ ݇ሻଶ ൌ ‫ ݎ‬ଶ

υχ

‫ݔ‬

ܲ ൌ ሺ‫ݕ ,ݔ‬ሻ

‫ݎ‬

‫ ܥ‬ൌ ሺ݄, ݇ሻ

MA1200 Basic Calculus and Linear Algebra I
Lecture Note 1: Coordinate Geometry and Conic Sections

Example 4

(a) Find the equation of the circle with centre ሺ1, െ1ሻ and radius √2.
(b) The equation of a circle is given by ሺ‫ ݔ‬൅ 3ሻଶ ൅ ሺ‫ ݕ‬െ 1ሻଶ ൌ 7. Find the centre and the radius of this circle,

☺Solution:
(a) The equation of the circle is



ሺ‫ ݔ‬െ 1ሻଶ ൅ ൫‫ ݕ‬െ ሺെ1ሻ൯ ൌ ൫√2൯ ⇒ ሺ‫ ݔ‬െ 1ሻଶ ൅ ሺ‫ ݕ‬൅ 1ሻଶ ൌ 2
⇒ ‫ ݔ‬ଶ ൅ ‫ ݕ‬ଶ െ 2‫ ݔ‬൅ 2‫ ݕ‬ൌ 0.

(b)



In order to find the centre and radius, one has to express the equation into the form ሺ‫ ݔ‬െ ݄ሻଶ ൅ ሺ‫ ݕ‬െ ݇ሻଶ ൌ ‫ ݎ‬ଶ .

Note that

ሻଶ

ሺ‫ ݔ‬൅ 3

ሻଶ

൅ ሺ‫ ݕ‬െ 1

ൌ 7 ⇒ ൫‫ ݔ‬െ ሺെ3ሻ൯ ൅ ሺ‫ ݕ‬െ 1


So the circle has centre ‫ ܥ‬ൌ ሺെ3,1ሻ and radius √7.

υψ

MA1200 Basic Calculus and Linear Algebra I
Lecture Note 1: Coordinate Geometry and Conic Sections

ሻଶ



ൌ ൫√7൯

Example 5 (Useful technique: Completing Square)

The equation of a circle is given by ‫ ݔ‬ଶ ൅ ‫ ݕ‬ଶ ൅ 8‫ ݔ‬െ 10‫ ݕ‬െ 8 ൌ 0. Find the centre and radius of this circle.
☺Solution:
Note that

‫ ݔ‬ଶ ൅ ‫ ݕ‬ଶ ൅ 8‫ ݔ‬െ 10‫ ݕ‬െ 8 ൌ 0 ⇒ ሺ‫ ݔ‬ଶ ൅ 8‫ݔ‬ሻ ൅ ሺ‫ ݕ‬ଶ െ 10‫ݕ‬ሻ െ 8 ൌ 0.
⇒ ሺ‫ ݔ‬ଶ ൅ 2ሺ4ሻ‫ݔ‬ሻ ൅ ሺ‫ ݕ‬ଶ െ 2ሺ5ሻ‫ݕ‬ሻ െ 8 ൌ 0

ሺ‫ ݔ‬ଶᇧ 2
⇒ ᇣᇧ ൅ᇧ ሺ4ሻ‫ ݔ‬ᇧ ᇧᇥ െ 4ଶ ൅ ሺ‫ ݕ‬ଶᇧ ᇧ ሺ5ሻ‫ ݕ‬ᇧ ᇧᇥ െ 5ଶ െ 8 ൌ 0
ᇧ ᇧᇤᇧ ൅ 4ଶ ሻ
ᇧᇧ
ᇣᇧ െ 2
ᇧ ᇧᇤᇧ ൅ 5ଶ ሻ
ᇧᇧ
௔మ ାଶ௔௕ା௕మ ୀሺ௔ା௕ሻమ

௔మ ିଶ௔௕ା௕మ ୀሺ௔ି௕ሻమ

⇒ ሺ‫ ݔ‬൅ 4ሻଶ ൅ ሺ‫ ݕ‬െ 5ሻଶ ൌ 8 ൅ 5ଶ ൅ 4ଶ ൌ 49

⇒ ൫‫ ݔ‬െ ሺെ4ሻ൯ ൅ ሺ‫ ݕ‬െ 5ሻଶ ൌ 7ଶ .


So the centre is ‫ ܥ‬ൌ ሺെ4,5ሻ and the radius is 7.

υω

MA1200 Basic Calculus and Linear Algebra I
Lecture Note 1: Coordinate Geometry and Conic Sections

Example 6

Find the equation of the circle centered at ‫ ܥ‬ൌ ሺ2,3ሻ and passing through the point ܲ ൌ ሺെ2,1ሻ.
☺Solution:

We need to find the radius of the circle ‫.ݎ‬

‫ ܥ‬ൌ ሺ2,3ሻ

From the figure on the right, we have

‫ ݎ‬ൌ ‫ ܲܥ‬ൌ ට൫2 െ ሺെ2ሻ൯ ൅ ሺ3 െ 1ሻଶ ൌ √20.


Then the equation of circle is then given by
ሻଶ

ሺ‫ ݔ‬െ 2

υϊ

ሻଶ

൅ ሺ‫ ݕ‬െ 3



ܲ ൌ ሺെ2,1ሻ

ൌ ൫√20൯ ⇒ ‫ ݔ‬ଶ ൅ ‫ ݕ‬ଶ െ 4‫ ݔ‬െ 6‫ ݕ‬െ 7 ൌ 0.

MA1200 Basic Calculus and Linear Algebra I
Lecture Note 1: Coordinate Geometry and Conic Sections

‫ݎ‬

Conic Section #2: Ellipses

Given two fixed points ‫ܨ‬ଵ and ‫ܨ‬ଶ (called foci) on the plane, the ellipses is the set of all points ܲ such that
ܲ‫ܨ‬ଵ ൅ ܲ‫ܨ‬ଶ ൌ ܿ‫ ݐ݊ܽݐݏ݊݋‬ൌ 2ܽ.

(Note: The purpose of choosing the constant to be 2ܽ (instead of ܽ) is to make the equation of ellipses looks “nicer”).
ܲ

‫ܨ‬ଵ

‫ܥ‬

‫ܨ‬ଶ

Here, ‫ ܥ‬is the “centre” of ellipses and is also the “mid-point” between two foci ‫ܨ‬ଵ , ‫ܨ‬ଶ .

υϋ

MA1200 Basic Calculus and Linear Algebra I
Lecture Note 1: Coordinate Geometry and Conic Sections

Equation of Ellipses

For simplicity, we consider the ellipses with foci ‫ܨ‬ଵ ൌ ሺെܿ, 0ሻ and ‫ܨ‬ଶ ൌ ሺܿ, 0ሻ.
‫ݕ‬

To derive the equation of this ellipses, note that

ܲ‫ܨ‬ଵ ൅ ܲ‫ܨ‬ଶ ൌ 2ܽ

⇒ ට൫‫ ݔ‬െ ሺെܿ ሻ൯ ൅ ሺ‫ ݕ‬െ 0ሻଶ ൅ ඥሺ‫ ݔ‬െ ܿ ሻଶ ൅ ሺ‫ ݕ‬െ 0ሻଶ ൌ 2ܽ
ᇣᇧᇧᇧᇧᇧᇧᇧᇤᇧᇧᇧᇧᇧᇧᇧᇥ ᇣᇧᇧᇧᇧᇧᇧᇤᇧᇧᇧᇧᇧᇧᇥ


௉ிభ

௉ிమ

⇒ ඥሺ‫ ݔ‬൅ ܿ ሻଶ ൅ ‫ ݕ‬ଶ ൌ 2ܽ െ ඥሺ‫ ݔ‬െ ܿ ሻଶ ൅ ‫ ݕ‬ଶ

ሺെܿ, 0ሻ 0

Taking square on both sides and expand the term, we get

‫ ݔ‬ଶ ൅ 2‫ ܿݔ‬൅ ܿ ଶ ൅ ‫ ݕ‬ଶ ൌ 4ܽଶ െ 4ܽඥሺ‫ ݔ‬െ ܿ ሻଶ ൅ ‫ ݕ‬ଶ ൅ ‫ ݔ‬ଶ െ 2‫ ܿݔ‬൅ ܿ ଶ ൅ ‫ ݕ‬ଶ
ᇣᇧᇧᇧᇧᇧᇤᇧᇧᇧᇧᇧᇥ
⇒ 4ܽඥሺ‫ ݔ‬െ ܿ ሻଶ ൅ ‫ ݕ‬ଶ ൌ 4ܽଶ െ 4‫ܿݔ‬

ሺ௫ି௖ ሻమ ା௬ మ

⇒ ܽඥሺ‫ ݔ‬െ ܿ ሻଶ ൅ ‫ ݕ‬ଶ ൌ ܽଶ െ ‫ܿݔ‬

υό

MA1200 Basic Calculus and Linear Algebra I
Lecture Note 1: Coordinate Geometry and Conic Sections

ܲ ൌ ሺ‫ݕ ,ݔ‬ሻ
ሺܿ, 0ሻ

‫ݔ‬

Taking square on both sides again, we get
ܽଶ ሾሺ‫ ݔ‬െ ܿ ሻଶ ൅ ‫ ݕ‬ଶ ሿ ൌ ܽସ െ 2ܽଶ ‫ ܿݔ‬൅ ‫ ݔ‬ଶ ܿ ଶ

Expanding the terms and rearranging them, we finally get
ሺܽଶ െ ܿ ଶ ሻ‫ ݔ‬ଶ ൅ ܽଶ ‫ ݕ‬ଶ ൌ ܽଶ ሺܽଶ െ ܿ ଶ ሻ.

We define ܾଶ ൌ ܽଶ െ ܿ ଶ ൐ 0, then
ܾଶ ‫ ݔ‬ଶ ൅ ܽଶ ‫ ݕ‬ଶ ൌ ܽଶ ܾଶ

ܾଶ ‫ ݔ‬ଶ ܽଶ ‫ ݕ‬ଶ ܽଶ ܾଶ ‫ ݔ‬ଶ ‫ ݕ‬ଶ
⇒ ଶ ଶ ൅ ଶ ଶ ൌ ଶ ଶ ⇒ ଶ ൅ ଶ ൌ 1.
ܽ ܾ
ܽ ܾ
ܽ ܾ
ܽ
ܾ

This is OK since
ܲ‫ܨ‬ଵ ൅ ܲ‫ܨ‬ଶ ൐ ‫ܨ‬ଵ ‫ܨ‬ଶ ⇒ 2ܽ ൐ 2ܿ
⇒ܽ൐ܿ

Equation of Ellipse (Standard Case)

For ܽ, ܿ ൐ 0, the equation of ellipse with foci ሺܿ, 0ሻ and ሺെܿ, 0ሻ is given by
‫ݔ‬ଶ ‫ݕ‬ଶ
൅ ଶ ൌ 1, ‫ܾ ݁ݎ݄݁ݓ‬ଶ ൌ ܽଶ െ ܿ ଶ , ܽ ൐ ܾ ൐ 0

ܽ
ܾ

Here, 2ܽ is the sum of distance from any point on ellipse to the two foci.

υύ

MA1200 Basic Calculus and Linear Algebra I
Lecture Note 1: Coordinate Geometry and Conic Sections

Some remarks on the ellipse ௔మ ൅ ௕మ ൌ 1
௫మ

௬మ

• The ellipse passes through the points ሺܽ, 0ሻ, ሺെܽ, 0ሻ,
ሺ0, ܾሻ, ሺ0, െܾሻ (they are called vertices).

• The ellipse is centered at the origin ሺ0,0ሻ.

ሺെܽ, 0ሻ

‫ݕ‬

ሺ0, ܾሻ
0

ሺܽ, 0ሻ

ሺ0, െܾሻ

‫ݔ‬

• If the foci ‫ܨ‬ଵ ൌ ሺ0, ܿሻ and ‫ܨ‬ଶ ൌ ሺ0, െܿሻ lies on ‫-ݕ‬axis instead of ‫-ݔ‬axis, the corresponding equation of the ellipse formed is seen to be ‫ݕ‬
ሺ0, ܽሻ
‫ݔ‬ଶ ‫ݕ‬ଶ
൅ ଶ ൌ 1,
‫ 2ܾ ݁ݎ݄݁ݓ‬ൌ ܽ2 െ ܿ2 .
‫ܨ‬ଵ ሺܾ, 0ሻ
ܾଶ ܽ
ሺെܾ, 0ሻ
‫ݔ‬
(The role of ܽ and ܾ are reversed)
0
• When ܽ ൌ ܾ, then the equation of ellipse becomes
మ ൅

௫మ



φτ

௬మ
௔మ

ൌ 1 ⇒ ‫ ݔ‬ଶ ൅ ‫ ݕ‬ଶ ൌ ܽଶ which is the equation of circle.

MA1200 Basic Calculus and Linear Algebra I
Lecture Note 1: Coordinate Geometry and Conic Sections

‫ܨ‬ଶ

ሺ0, െܽሻ

Example 7

The equation of an ellipse is 4‫ ݔ‬ଶ ൅ 9‫ ݕ‬ଶ ൌ 36. Sketch the graph. Find the coordinates of vertices and the foci of this ellipse.
☺Solution:

4‫ ݔ‬ଶ ൅ 9‫ ݕ‬ଶ
‫ݔ‬ଶ ‫ݕ‬ଶ
‫ݔ‬ଶ ‫ݕ‬ଶ
4‫ ݔ‬ଶ ൅ 9‫ ݕ‬ଶ ൌ 36 ⇒
ൌ1⇒

ൌ1⇒ ଶ൅ ଶൌ1
36
9
4
3
2
The ellipse passes through the vertices ሺ3,0ሻ, ሺെ3,0ሻ, ሺ0,2ሻ, ሺ0, െ2ሻ.

From the graph of ellipse, we see the foci lies on ‫-ݔ‬axis and

ܿ ൌ 3 െ 2 ⇒ ܿ ൌ √5.






Hence the foci of this ellipse is

‫ܨ‬ଵ ൌ ൫െ√5, 0൯ and ‫ܨ‬ଶ ൌ ൫√5, 0൯.

φυ

ሺെ3,0ሻ

‫ݕ‬

ሺ0,2ሻ

0

ሺ0, െ2ሻ

MA1200 Basic Calculus and Linear Algebra I
Lecture Note 1: Coordinate Geometry and Conic Sections

ሺ3,0ሻ

‫ݔ‬

General equation of ellipse

In general, if the centre of the ellipse is ‫ ܥ‬ൌ ሺ݄, ݇ሻ instead of origin ሺ0,0ሻ, then the equation of ellipse is given by
Equation of Ellipse (General Case)

ሺ‫ ݔ‬െ ݄ሻଶ ሺ‫ ݕ‬െ ݇ሻଶ

ൌ1
ܽଶ
ܾଶ

where ‫ ܥ‬ൌ ሺ݄, ݇ሻ is the centre of the ellipse
ሺെܽ, 0ሻ

φφ

‫ݕ‬

ሺ0, ܾሻ
0

ሺ0, െܾሻ

ሺܽ, 0ሻ

‫ݔ‬

‫ݕ‬

ሺെܽ ൅ ݄, ݇ሻ
0

ሺ݄, ܾ ൅ ݇ሻ
ሺ݄, ݇ሻ

ሺܽ ൅ ݄, ݇ሻ

ሺ݄, െܾ ൅ ݇ሻ

݄ units

MA1200 Basic Calculus and Linear Algebra I
Lecture Note 1: Coordinate Geometry and Conic Sections

‫ݔ‬

݇ units

Example 8

The equation of the ellipse is 4‫ ݔ‬ଶ ൅ 9‫ ݕ‬ଶ െ 8‫ ݔ‬൅ 36‫ ݕ‬െ 9 ൌ 0. Sketch the graph of this ellipse and indicate the coordinates of the vertexes.
☺Solution:
Similar to Example 5, we rewrite the equation into the form
1 using completing square technique

4‫ ݔ‬ଶ ൅ 9‫ ݕ‬ଶ െ 8‫ ݔ‬൅ 36‫ ݕ‬െ 9 ൌ 0

ሺ௫ି௛ሻమ
௔మ

⇒ 4ሺ‫ ݔ‬ଶ െ 2‫ ݔ‬ሻ ൅ 9ሺ‫ ݕ‬ଶ ൅ 4‫ݕ‬ሻ െ 9 ൌ 0

⇒ 4ሺ‫ ݔ‬ଶ െ 2ሺ1ሻ‫ ݔ‬൅ 1ଶ െ 1ଶ ሻ ൅ 9ሺ‫ ݕ‬ଶ ൅ 2ሺ2ሻ‫ ݕ‬൅ 2ଶ െ 2ଶ ሻ െ 9 ൌ 0
⇒ 4ሺ‫ ݔ‬െ 1ሻଶ ൅ 9ሺ‫ ݕ‬൅ 2ሻଶ ൌ 49



ሺ‫ ݔ‬െ 1ሻଶ
7 ଶ
ቀ2ቁ

φχ



൫‫ ݕ‬െ ሺെ2ሻ൯
7 ଶ
ቀ3ቁ



ൌ 1.

MA1200 Basic Calculus and Linear Algebra I
Lecture Note 1: Coordinate Geometry and Conic Sections



ሺ௬ି௞ሻమ
௕మ



The graph of this ellipse is sketched below:

7
1
൬1, െ2 ൅ ൰ ൌ ൬1, ൰
3
3

7
൬1 െ , െ2൰
2

5
ൌ ൬െ , െ2൰
2

ሺ1, െ2ሻ

7
൬1 ൅ , െ2൰
2
9
ൌ ൬ , െ2൰
2

7
13
൬1, െ2 െ ൰ ൌ ൬1, െ ൰
3
3

The coordinates of the vertexes are given by

1
13
5
9
൬1, ൰ , ൬1, െ ൰ , ൬െ , െ2൰ ܽ݊݀ ൬ , െ2൰.
3
3
2
2

φψ

MA1200 Basic Calculus and Linear Algebra I
Lecture Note 1: Coordinate Geometry and Conic Sections

Conic Section #3: Parabola
A parabola is the set of all points in a plane that are equidistant from a fixed line (called directix) and a fixed point (called focus).

Parabola
Axis of symmetry Fixed points
(Focus)

Fixed lines
(Directix)
Vertex

φω

MA1200 Basic Calculus and Linear Algebra I
Lecture Note 1: Coordinate Geometry and Conic Sections

Equation of parabola (Standard Case)

We consider the parabola with focus at ‫ ܨ‬ൌ ሺ0, ܽሻ and the directix ‫ ݕ‬ൌ െܽ,
‫ݕ‬
where ܽ ൐ 0.
Let ܲ ൌ ሺ‫ݕ ,ݔ‬ሻ be any point on the parabola.
According to the definition, we must have
ܲܳ ൌ ܲ‫ܨ‬

ܲ ൌ ሺ‫ݕ ,ݔ‬ሻ

⇒ ‫ ݕ‬െ ሺെܽሻ ൌ ඥሺ‫ ݔ‬െ 0ሻଶ ൅ ሺ‫ ݕ‬െ ܽሻଶ
⇒ ሺ‫ ݕ‬൅ ܽሻଶ ൌ ‫ ݔ‬ଶ ൅ ሺ‫ ݕ‬െ ܽሻଶ

ܳ ൌ ሺ‫ ,ݔ‬െܽሻ

⇒ ‫ ݕ‬ଶ ൅ 2ܽ‫ ݕ‬൅ ܽଶ ൌ ‫ ݔ‬ଶ ൅ ‫ ݕ‬ଶ െ 2ܽ‫ ݕ‬൅ ܽଶ ⇒ ‫ ݔ‬ଶ ൌ 4ܽ‫.ݕ‬

0

Equation of parabola (Standard Case)

‫ ܨ‬ൌ ሺ0, ܽሻ

‫ݔ‬

‫ ݕ‬ൌ െܽ

For ܽ ൐ 0, the equation of parabola with focus ሺ0, ܽሻ and directix ‫ ݕ‬ൌ െܽ is given by
‫ ݔ‬ଶ ൌ 4ܽ‫.ݕ‬

φϊ

MA1200 Basic Calculus and Linear Algebra I
Lecture Note 1: Coordinate Geometry and Conic Sections

Remark on standard equation of parabola

• The vertex of the parabola is simply the origin ܱ ൌ ሺ0,0ሻ and the ‫-ݕ‬axis is axis of symmetry.
• One can obtain different parabola with different focus and the directix, some common types of parabola are shown below:
‫ݕ‬
‫ݕ‬

‫ ݔ‬ൌ െܽ

ܱ
‫ ܨ‬ൌ ሺ0, െܽሻ

‫ ݔ‬ଶ ൌ െ4ܽ‫ݕ‬
(ܽ becomes െܽ)

φϋ

‫ݕ‬

‫ݔ‬

ܱ

‫ݔ‬ൌܽ

‫ ܨ‬ൌ ሺܽ, 0ሻ

‫ ݕ‬ଶ ൌ 4ܽ‫ݔ‬
(Interchange ‫ݔ‬ and ‫)ݕ‬

‫ݔ‬

‫ ܨ‬ൌ ሺെܽ, 0ሻ

ܱ

‫ ݕ‬ଶ ൌ െ4ܽ‫ݔ‬

MA1200 Basic Calculus and Linear Algebra I
Lecture Note 1: Coordinate Geometry and Conic Sections

‫ݔ‬

Example 9

The vertex and the axis of symmetry of parabola are the origin and the ‫-ݔ‬axis respectively. If the parabola passes through the point ሺ6,3ሻ. Find the equation of this parabola.
☺Solution:

Note that the parabola is symmetric about ‫-ݔ‬axis and has vertex at the origin, the equation of parabola is of the form ‫ ݕ‬ଶ ൌ 4ܽ‫.ݔ‬

Since the parabola passes through the point ሺ6,3ሻ, we substitute ሺ‫ݕ ,ݔ‬ሻ ൌ ሺ6,3ሻ into the equation and get
3
3 ൌ 4ܽሺ6ሻ ⇒ ܽ ൌ .
8
Then the equation of parabola is given by


φό

3
3
‫ ݕ‬ଶ ൌ 4 ൬ ൰ ‫ ݕ ⇒ ݔ‬ଶ ൌ ‫.ݔ‬
8
2

MA1200 Basic Calculus and Linear Algebra I
Lecture Note 1: Coordinate Geometry and Conic Sections

More general equation of parabola

In general, the graph of a parabola can have the vertex ሺ݄, ݇ሻ (instead of origin
ሺ0,0ሻ). Depending on the axis of symmetry, one can derive the following general equation of parabola for this case:
Equation of Parabola (More General Case)

Take ܽ ് 0, the equation of parabola with vertex ሺ݄, ݇ሻ is given by
(i) If the axis of symmetry is ‫ ݕ‬ൌ ݇,

ሺ‫ ݔ‬െ ݄ሻଶ ൌ 4ܽሺ‫ ݕ‬െ ݇ሻ.

(ii) If the axis of symmetry is ‫ ݔ‬ൌ ݄,

ሺ‫ ݕ‬െ ݇ሻଶ ൌ 4ܽሺ‫ ݔ‬െ ݄ሻ.

Roughly speaking, the equation in general case can be obtained by replacing ‫ݔ‬ with ‫ ݔ‬െ ݄ and ‫ ݕ‬with ‫ ݕ‬െ ݇ in the standard form of the equation.

φύ

MA1200 Basic Calculus and Linear Algebra I
Lecture Note 1: Coordinate Geometry and Conic Sections

Conic Section #4: Hyperbola

Let ‫ܨ‬ଵ and ‫ܨ‬ଶ be two fixed points (foci) in the plane, a hyperbola is the set of all points ܲ in a plane such that
|ܲ‫ܨ‬ଵ െ ܲ‫ܨ‬ଶ | ൌ ܿ‫ ݐ݊ܽݐݏ݊݋‬ൌ 2ܽ

i.e. the difference between ܲ‫ܨ‬ଵ and ܲ‫ܨ‬ଶ is always fixed.
‫ݕ‬

‫ܨ‬ଵ

χτ

0

ܲ
‫ܨ‬ଶ

‫ݔ‬

MA1200 Basic Calculus and Linear Algebra I
Lecture Note 1: Coordinate Geometry and Conic Sections

Equation of Hyperbola (Standard Form)

For simplicity, we consider the case when the foci are ‫ܨ‬ଵ ൌ ሺെܿ, 0ሻ and
‫ܨ‬ଶ ൌ ሺܿ, 0ሻ.
‫ݕ‬

ܲ ൌ ሺ‫ݕ ,ݔ‬ሻ

According to the definition of hyperbola:

|ܲ‫ܨ‬ଵ െ ܲ‫ܨ‬ଶ | ൌ 2ܽ

⇒ ሺܲ‫ܨ‬ଵ െ ܲ‫ܨ‬ଶ ሻଶ ൌ ሺ2ܽሻଶ ൌ 4ܽଶ



‫ܨ‬ଵ ൌ ሺെܿ, 0ሻ

⇒ ቈට൫‫ ݔ‬െ ሺെܿ ሻ൯ ൅ ‫ ݕ‬ଶ െ ඥሺ‫ ݔ‬െ ܿ ሻଶ ൅ ‫ ݕ‬ଶ ቉ ൌ 4ܽଶ


After a tedious calculation, one can obtain
ሺܿ ଶ െ ܽଶ ሻ‫ ݔ‬ଶ െ ܽଶ ‫ ݕ‬ଶ ൌ ܽଶ ሺܿ ଶ െ ܽଶ ሻ

We let ܾ ଶ ൌ ܿ ଶ െ ܽଶ ൐ 0 (ܿ ൐ ܽ, why??), then
‫ݔ‬ଶ ‫ݕ‬ଶ
ܾ ‫ ݔ‬െ ܽ ‫ ݕ‬ൌ ܽ ܾ ⇒ ଶ െ ଶ ൌ 1.
ܽ
ܾ
ଶ ଶ

χυ

ଶ ଶ

ଶ ଶ

MA1200 Basic Calculus and Linear Algebra I
Lecture Note 1: Coordinate Geometry and Conic Sections

0

‫ܨ‬ଶ ൌ ሺܿ, 0ሻ
‫ݔ‬
centre

Equation of Hyperbola (Standard Case)

For ܿ ൐ 0, the equation of hyperbola with foci ሺܿ, 0ሻ and ሺെܿ, 0ሻ is given by
‫ݔ‬ଶ ‫ݕ‬ଶ
െ ଶ ൌ 1, ‫ܾ ݁ݎ݄݁ݓ‬ଶ ൌ ܿ ଶ െ ܽଶ ,

ܽ
ܾ

ܽ, ܾ ൐ 0

Here, 2ܽ is the difference between distance from any point on ellipse to the two foci.

Remark on standard equation of parabola

• The origin (mid-point of foci ‫ܨ‬ଵ and ‫ܨ‬ଶ ) is the centre of the hyperbola.
• ‫ܣ‬ଵ ൌ ሺെܽ, 0ሻ, ‫ܣ‬ଶ ൌ ሺܽ, 0ሻ are called vertices of the hyperbola.
• When ‫ ݕ ,ݔ‬get larger, the graph gets close

ܾ
‫ݕ‬ൌെ ‫ݔ‬
ܽ

to the pair of straight lines ‫ ݕ‬ൌ ‫ ݔ‬and




‫ ݕ‬ൌ െ ‫ ݔ‬which are called asymptotes.


χφ



MA1200 Basic Calculus and Linear Algebra I
Lecture Note 1: Coordinate Geometry and Conic Sections

‫ݕ‬ൌ

ܾ
‫ݔ‬
ܽ

• If the foci are ‫ܨ‬ଵ ൌ ሺ0, ܿሻ and ‫ܨ‬ଶ ൌ ሺ0, െܿሻ, then the corresponding equation of the hyperbola is given by
‫ݕ‬ଶ ‫ݔ‬ଶ
െ ଶ ൌ 1,
ܾ ଶ ൌ ܿ ଶ െ ܽଶ .

ܽ
ܾ
The centre remains to be the origin and the vertices are ‫ܣ‬ଵ ൌ ሺ0, െܽሻ,
‫ܣ‬ଶ ൌ ሺ0, ܽሻ.
ܾ
‫ݔ‬ൌെ ‫ݕ‬
ܽ

χχ

ܾ
‫ݔ‬ൌ ‫ݕ‬
ܽ

MA1200 Basic Calculus and Linear Algebra I
Lecture Note 1: Coordinate Geometry and Conic Sections

• In general, the equation of hyperbola is of the following general form.
Equation of Hyperbola (General Form)

ሺ‫ ݔ‬െ ݄ሻଶ ሺ‫ ݕ‬െ ݇ሻଶ

ൌ 1,
ܽଶ
ܾଶ
ሺ‫ ݕ‬െ ݇ሻଶ ሺ‫ ݔ‬െ ݄ሻଶ

ൌ1
ܽଶ
ܾଶ

where the centre is ‫ ܥ‬ൌ ሺ݄, ݇ሻ.
Example 10

It is given that the equation of a conic section is െ‫ ݔ‬ଶ ൅ 4‫ ݕ‬ଶ ൅ 4‫ ݔ‬െ 24‫ ݕ‬൅
28 ൌ 0. Show that the conic section is hyperbola. Find the coordinates of the centre and the vertices of the hyperbola.

χψ

MA1200 Basic Calculus and Linear Algebra I
Lecture Note 1: Coordinate Geometry and Conic Sections

☺Solution:

െ‫ ݔ‬ଶ ൅ 4‫ ݕ‬ଶ ൅ 4‫ ݔ‬െ 24‫ ݕ‬൅ 28 ൌ 0

⇒ െሺ‫ ݔ‬ଶ െ 4‫ݔ‬ሻ ൅ 4ሺ‫ ݕ‬ଶ െ 6‫ݕ‬ሻ ൅ 28 ൌ 0

⇒ െሺ‫ ݔ‬ଶ െ 2ሺ2ሻ‫ ݔ‬൅ 2ଶ െ 2ଶ ሻ ൅ 4ሺ‫ ݕ‬ଶ െ 2ሺ3ሻ‫ ݕ‬൅ 3ଶ െ 3ଶ ሻ ൅ 28 ൌ 0
⇒ െሺ‫ ݔ‬െ 2ሻଶ ൅ 4ሺ‫ ݕ‬െ 3ሻଶ െ 4 ൌ 0

ሺ‫ ݔ‬െ 2ሻଶ
⇒ ሺ‫ ݕ‬െ 3ሻଶ െ
ൌ1
4

ሺ‫ ݕ‬െ 3ሻଶ ሺ‫ ݔ‬െ 2ሻଶ


ൌ 1.


1
2

So the conic section is a hyperbola.
The centre of hyperbola is ሺ2,3ሻ.

ܸଵ ൌ ሺ2,3 ൅ ܽሻ
ሺ2,3ሻ
ܸଶ ൌ ሺ2,3 െ ܽሻ

With ܽ ൌ 1, the coordinates of the vertices are ൫2, ሺ3 ൅ 1ሻ൯ ൌ ሺ2,4ሻ and
ሺ2, ሺ3 െ 1ሻሻ ൌ ሺ2,2ሻ.

χω

MA1200 Basic Calculus and Linear Algebra I
Lecture Note 1: Coordinate Geometry and Conic Sections

Classification of conic section
Recall that the equations of the above four conic sections are given by
• Circle -- ሺ‫ ݔ‬െ ݄ሻଶ ൅ ሺ‫ ݕ‬െ ݇ሻଶ ൌ ‫ ݎ‬ଶ ,
• Ellipse --

ሺ௫ି௛ሻమ
௔మ



ሺ௬ି௞ ሻమ
௕మ


ൌ 1,

• Parabola -- ሺ‫ ݔ‬െ ݄ሻ ൌ 4ܽሺ‫ ݕ‬െ ݇ሻ or ሺ‫ ݕ‬െ ݇ሻଶ ൌ 4ܽሺ‫ ݔ‬െ ݄ሻ,
• Hyperbola --

ሺ௫ି௛ሻమ
௔మ



ሺ௬ି௞ ሻమ
௕మ

ൌ 1,

ሺ௬ି௞ ሻమ
௔మ



ሺ௫ି௛ሻమ
௕మ

ൌ 1.

By expanding the terms, one can see that each of the equations can be expressed in the following form:
‫ ݔܣ‬ଶ ൅ ‫ ݕܥ‬ଶ ൅ ‫ ݔܦ‬൅ ‫ ݕܧ‬൅ ‫ ܨ‬ൌ 0 … … ሺ∗ሻ

where ‫ ܧ ,ܦ ,ܥ ,ܣ‬and ‫ ܨ‬are some numbers.

χϊ

MA1200 Basic Calculus and Linear Algebra I
Lecture Note 1: Coordinate Geometry and Conic Sections

In many cases, one may be only given the equation of the conic section of the form ሺ∗ሻ. In order to classify it (circle, ellipse, parabola or hyperbola), one can adopt “completing square technique” [see Example 5, 8, 10] to rewrite the equation into the “standard form”.
Example 11
Classify the type of conic section described by each of the following equations.
(a) 4‫ ݔ‬ଶ െ 16‫ ݔ‬൅ 25‫ ݕ‬ଶ െ 84 ൌ 0
(b) 4‫ ݔ‬ଶ ൅ 4‫ ݕ‬ଶ ൅ 8‫ ݔ‬െ 24‫ ݕ‬൅ 15 ൌ 0

☺Solution:

(a) Note that
4‫ ݔ‬ଶ െ 16‫ ݔ‬൅ 25‫ ݕ‬ଶ െ 84 ൌ 0
⇒ 4ሺ‫ ݔ‬ଶ െ 4‫ݔ‬ሻ ൅ 25‫ ݕ‬ଶ െ 84 ൌ 0
⇒ 4ሺ‫ ݔ‬ଶ െ 2ሺ2ሻ‫ ݔ‬൅ 2ଶ െ 2ଶ ሻ ൅ 25‫ ݕ‬ଶ െ 84 ൌ 0
⇒ 4ሺ‫ ݔ‬െ 2ሻଶ ൅ 25‫ ݕ‬ଶ ൌ 100

χϋ

MA1200 Basic Calculus and Linear Algebra I
Lecture Note 1: Coordinate Geometry and Conic Sections

ሺ‫ ݔ‬െ 2ሻଶ ‫ ݕ‬ଶ
ሺ‫ ݔ‬െ 2ሻଶ ‫ ݕ‬ଶ


ൌ1⇒
൅ ଶ ൌ 1.

25
4
5
2
So this equation represents an ellipse.
(b) Note that
4‫ ݔ‬ଶ ൅ 4‫ ݕ‬ଶ ൅ 8‫ ݔ‬െ 24‫ ݕ‬൅ 15 ൌ 0
⇒ 4ሺ‫ ݔ‬ଶ ൅ 2‫ݔ‬ሻ ൅ 4ሺ‫ ݕ‬ଶ െ 6‫ݕ‬ሻ ൅ 15 ൌ 0
⇒ 4ሺ‫ ݔ‬ଶ ൅ 2ሺ1ሻ‫ ݔ‬൅ 1ଶ െ 1ଶ ሻ ൅ 4ሺ‫ ݕ‬ଶ െ 2ሺ3ሻ‫ ݕ‬൅ 3ଶ െ 3ଶ ሻ ൅ 15 ൌ 0
⇒ 4ሺ‫ ݔ‬൅ 1ሻଶ ൅ 4ሺ‫ ݕ‬െ 3ሻଶ ൌ 25
25


⇒ ሺ‫ ݔ‬൅ 1ሻ ൅ ሺ‫ ݕ‬െ 3ሻ ൌ
4
5 ଶ

⇒ ൫‫ ݔ‬െ ሺെ1ሻ൯ ൅ ሺ‫ ݕ‬െ 3ሻଶ ൌ ൬ ൰
2
So the equation represents a circle.

χό

MA1200 Basic Calculus and Linear Algebra I
Lecture Note 1: Coordinate Geometry and Conic Sections

In general, given the equation of conic section ‫ ݔܣ‬ଶ ൅ ‫ ݕܥ‬ଶ ൅ ‫ ݔܦ‬൅ ‫ ݕܧ‬൅ ‫ ܨ‬ൌ 0, one can adopt the similar technique to classify the conic section. The flow chart below summarizes the result: (assuming at least one of ‫ ܥ ,ܣ‬is non-zero)
Equation of Conic Section

‫ ܥܣ‬ൌ 0

Parabola

‫ ݔܣ‬ଶ ൅ ‫ ݕܥ‬ଶ ൅ ‫ ݔܦ‬൅ ‫ ݕܧ‬൅ ‫ ܨ‬ൌ 0
‫ ܥܣ‬൏ 0
Hyperbola

‫ ܥܣ‬൐ 0

Circle/ Ellipse
‫ܣ‬ൌ‫ܥ‬
Circle

χύ

‫ܥ്ܣ‬
Ellipse

MA1200 Basic Calculus and Linear Algebra I
Lecture Note 1: Coordinate Geometry and Conic Sections

Rough reason of the above result:

• If ‫ ܥܣ‬ൌ 0, then either ‫ ܣ‬ൌ 0 or ‫ ܥ‬ൌ 0.
So the equation must be of the forms
‫ ݕܥ‬ଶ ൅ ‫ ݔܦ‬൅ ‫ ݕܧ‬൅ ‫ ܨ‬ൌ 0 ሺ‫0 ് ܦ‬ሻ ‫ ݔܣ ݎ݋‬ଶ ൅ ‫ ݔܦ‬൅ ‫ ݕܧ‬൅ ‫ ܨ‬ൌ 0 ሺ‫0 ് ܧ‬ሻ
Using completing square, one can rewrite the equations as
ሺ‫ݕ‬െ? ሻଶ ൌ 4ܽሺ‫ݔ‬െ? ሻ ‫ ݎ݋‬ሺ‫ݔ‬െ? ሻଶ ൌ 4ܽሺ‫ݕ‬െ? ሻ which are equation of parabola.
Remark:
In the unlikely case which ‫ ܦ‬ൌ 0 (for 1st equation) or ‫ ܧ‬ൌ 0 (for two equations). Then the 1st equation become
‫ ݕܥ‬ଶ ൅ ‫ ݕܧ‬൅ ‫ ܨ‬ൌ 0 … … ሺ∗∗ሻ
Depending on the number of roots of equation ሺ∗∗ሻ, the conic section will be
‫ ݕ‬ൌ ‫ݕ‬ଵ , ‫ ݕ‬ൌ ‫ݕ‬ଶ ݂݅ ሺ∗∗ሻ ݄ܽ‫ݏ݊݋݅ݐݑ݈݋ݏ ݋ݓݐ ݏ‬
ቐ ‫ ݕ‬ൌ ‫ݕ‬଴ ݂݅ ሺ∗∗ሻ ݄ܽ‫. ݊݋݅ݐݑ݈݋ݏ ݁݊݋ ݏ‬
݊‫ ݂݅ ݈݁݊݅ ݋‬ሺ∗∗ሻ ݄ܽ‫݊݋݅ݐݑ݈݋ݏ ݋݊ ݏ‬

ψτ

MA1200 Basic Calculus and Linear Algebra I
Lecture Note 1: Coordinate Geometry and Conic Sections

• If ‫ ܥܣ‬൏ 0, then either ‫ ܣ‬൐ 0, ‫ ܥ‬൏ 0 or ‫ ܣ‬൏ 0, ‫ ܥ‬൐ 0.
Then after completing square, the equation can be rewritten as
ሺ‫ݔ‬െ? ሻଶ ሺ‫ݕ‬െ? ሻଶ
‫ۓ‬

ൌ1


?
?
‫ܣ‬ሺ‫ݔ‬െ? ሻଶ ൅ ‫ ܥ‬ሺ‫ݕ‬െ? ሻଶ ൌ ‫ ܨ‬ᇱ ⇒ ‫ ݁ݎ݄݁ݓ‬ሺ‫ ܨ‬ᇱ ് 0ሻ

ሺ‫ݔ‬െ? ሻଶ
‫۔‬ሺ‫ݕ‬െ? ሻ
‫? ە‬ଶ െ ?ଶ ൌ 1 which is equation of hyperbola.
Remark: (Degenerate Case)
In the unlikely case when ‫ ܨ‬ᇱ ൌ 0, the equation becomes

‫ܣ‬
‫ۓ‬
ඨ ‫ݔ‬൅?
ۖ‫ݕ‬ൌ ‫ܥ‬

‫ܣ‬
‫ܣ‬ሺ‫ݔ‬െ? ሻଶ ൅ ‫ ܥ‬ሺ‫ݕ‬െ? ሻଶ ൌ 0 ⇒ ‫ݕ‬െ? ൌ േඨെ ሺ‫ݔ‬െ? ሻ ⇒

‫ܥ‬
‫۔‬
‫ܣ‬
வ଴
ۖ‫ ݕ‬ൌ െඨ ‫ݔ‬൅?
‫ܥ‬
‫ە‬ which represents a pair of (unparallel) straight lines.

ψυ

MA1200 Basic Calculus and Linear Algebra I
Lecture Note 1: Coordinate Geometry and Conic Sections

• If ‫ ܥܣ‬൐ 0, then ‫ ܥ ,ܣ‬have the same sign (both positive or both negative).
After completing square, the equation can be rewritten as
ሺ‫ݔ‬െ? ሻଶ ሺ‫ݕ‬െ? ሻଶ
ሺ ᇱ

‫ܣ‬ሺ‫ݔ‬െ? ሻଶ ൅ ‫ ܥ‬ሺ‫ݕ‬െ? ሻଶ ൌ ‫ ܨ‬ᇱ ⇒
ଶ൅
ଶ ൌ 1, ‫0 ് ܨ ݁ݎ݄݁ݓ‬


ට‫ ܨ‬ቇ
ට‫ ܨ‬ቇ


‫ܥ‬
‫ܣ‬ which is the equation of circle or ellipse.
If ‫ ܣ‬ൌ ‫ ,ܥ‬then

ிᇲ




ிᇲ


ൌ ‫ ݎ‬ଶ . Hence the equation can be expressed as

ሺ‫ݔ‬െ? ሻଶ ൅ ሺ‫ݕ‬െ? ሻଶ ൌ ‫ ݎ‬ଶ which is the equation of circle.

If ‫ ,ܥ ് ܣ‬then

ellipse (

ሺ௫ି௛ሻమ
௔మ

ிᇲ




Remark: (Degenerate Case)



ிᇲ


ሺ௬ି௞ ሻమ
௕మ

. Then the equation is simply the equation of
ൌ 1).

In the unlikely case when ‫ ܨ‬ᇱ ൌ 0, then the conic section will become a single point.

ψφ

MA1200 Basic Calculus and Linear Algebra I
Lecture Note 1: Coordinate Geometry and Conic Sections

Example 12
Identify the graph of the following functions
(a) 3‫ ݔ‬ଶ െ 2‫ ݕ‬ଶ ൅ 5‫ ݔ‬െ ‫ ݕ‬െ 5 ൌ 0
(b) 2‫ ݔ‬ଶ ൅ 2‫ ݕ‬ଶ െ ‫ ݔ‬൅ ‫ ݕ‬െ 7 ൌ 0
(c) ‫ ݕ‬ଶ െ 4‫ ݔ‬൅ 2‫ ݕ‬െ 1 ൌ 0

☺Solution:

(a) Note that ‫ ܥܣ‬ൌ 3 ൈ ሺെ2ሻ ൌ െ6 ൏ 0, the graph is a hyperbola.

(b) Note that ‫ ܥܣ‬ൌ 2 ൈ 2 ൌ 4 ൐ 0, the graph is either circle or ellipse. Since
‫ ܣ‬ൌ ‫ ܥ‬ൌ 2, we conclude that the graph is a circle.

(c) Note that ‫ ܥܣ‬ൌ 0 ൈ 1 ൌ 0 and coefficient of ‫( ݔ‬ൌ െ4) is non-zero, the graph is a parabola.

ψχ

MA1200 Basic Calculus and Linear Algebra I
Lecture Note 1: Coordinate Geometry and Conic Sections

General Conic Section
In general, a conic section in 2-D can be expressed in the following form
‫ ݔܣ‬ଶ ൅ ‫ ݕݔܤ‬൅ ‫ ݕܥ‬ଶ ൅ ‫ ݔܦ‬൅ ‫ ݕܧ‬൅ ‫ ܨ‬ൌ 0

where ‫ ܧ ,ܦ ,ܥ ,ܤ ,ܣ‬and ‫ ܨ‬are constants.

• It is impossible for us to rewrite the equation into the form
ሺ‫ ݔ‬െ ݄ሻଶ ሺ‫ ݕ‬െ ݇ሻଶ

ൌ1
‫݌‬
‫ݍ‬ because of the existence of the term ‫.ݕݔܤ‬

• The term ‫ ݕݔܤ‬exists because the conic section (circle, ellipse, parabola or hyperbola) is being rotated from its “standard position”.

ψψ

MA1200 Basic Calculus and Linear Algebra I
Lecture Note 1: Coordinate Geometry and Conic Sections

As an example, we suppose that the equation of a conic section is given by
13‫ ݔ‬ଶ ൅ 10‫ ݕݔ‬൅ 13‫ ݕ‬ଶ ൌ 72. We may try to identify the conic section by first sketching the graph of this conic section.
‫ݕ‬

The figure below shows the graph of the conic section

0

‫ݔ‬

One can guess that the conic section is an ellipse since one can see the
“standard ellipse” by rotating our “view point” by certain degree.

ψω

MA1200 Basic Calculus and Linear Algebra I
Lecture Note 1: Coordinate Geometry and Conic Sections

Mathematically, rotating our “view point” is equivalent to rotating the ‫-ݕݔ‬axis
(change to ‫ ݔ‬ᇱ ‫ ݕ‬ᇱ -axis).
‫ݕ‬

‫ݕ‬

‫ݔ‬
13‫ ݔ‬൅ 10‫ ݕݔ‬൅ 13‫ ݕ‬ൌ 72


45°

Rotating the axis by
45° in clockwise direction ଶ

(‫-ݕݔ‬coordinate)

8ሺ‫ ݔ‬ᇱ ሻଶ ൅ 18ሺ‫ ݕ‬ᇱ ሻଶ ൌ 72
ሺ‫ ݔ‬ᇱ ሻଶ ሺ‫ ݕ‬ᇱ ሻଶ
⇒ ଶ ൅ ଶ ൌ1
3
2
(‫-′ݕ′ݔ‬coordinate)

This technique is called “Rotation of Axes”.

ψϊ

MA1200 Basic Calculus and Linear Algebra I
Lecture Note 1: Coordinate Geometry and Conic Sections

‫ݔ‬

Rotation of Axes

Suppose the ‫-ݔ‬axis and ‫-ݕ‬axis are rotated through a “positive” angle (anticlockwise direction), we obtain a new ‫-’ݔ‬axis and ‫-′ݕ‬axis and this forms a ‫′ݕ′ݔ‬coordinate system:
‫ݕ‬

ܲ
‫ݎ‬
ߙ

ߠ

‫ݔ‬

In ሺ‫ݕ ,ݔ‬ሻ-coordinate, the coordinate of ܲ is ‫ ݔ‬ൌ ‫ ݎ‬cosሺߠ ൅ ߙሻ , ‫ ݕ‬ൌ ‫ ݎ‬sinሺߠ ൅ ߙሻ.

In ሺ‫ ݔ‬ᇱ , ‫ ݕ‬ᇱ ሻ-coordinate, the coordinate of ܲ is ‫ ݔ‬ᇱ ൌ ‫ ݎ‬cos ߙ, ‫ ݕ‬ᇱ ൌ ‫ ݎ‬sin ߙ.

ψϋ

MA1200 Basic Calculus and Linear Algebra I
Lecture Note 1: Coordinate Geometry and Conic Sections

Using the compound angle formula (details will be discussed in later Chapter), cosሺߙ ൅ ߠሻ ൌ cos ߙ cos ߠ െ sin ߙ sin ߠ,

we have

sinሺߙ ൅ ߠሻ ൌ sin ߙ cos ߠ ൅ cos ߙ sin ߠ,

‫ ݔ‬ൌ ‫ ݎ‬cosሺߙ ൅ ߠሻ ൌ ‫ ݎ‬cos ߙ cos ߠ െ ‫ ߙ ݊݅ݏݎ‬sin ߠ ൌ ‫ ݔ‬ᇱ cos ߠ െ ‫ ݕ‬ᇱ sin ߠ,

‫ ݕ‬ൌ ‫ ݎ‬sinሺߙ ൅ ߠሻ ൌ ‫ ݎ‬sin ߙ cos ߠ ൅ ‫ ݎ‬cos ߙ sin ߠ ൌ ‫ ݕ‬ᇱ cos ߠ ൅ ‫ ݔ‬ᇱ sin ߠ.

Summing up, we have the following relation between ‫-ݕݔ‬coordinate and ‫′ݕ′ݔ‬coordinate:
‫ ݔ‬ൌ ‫ ݔ‬ᇱ cos ߠ െ ‫ ݕ‬ᇱ sin ߠ,
‫ ݕ‬ൌ ‫ ݕ‬ᇱ cos ߠ ൅ ‫ ݔ‬ᇱ sin ߠ

ψό

MA1200 Basic Calculus and Linear Algebra I
Lecture Note 1: Coordinate Geometry and Conic Sections

Example 13

It is given that the equation of a conic section is ‫ ݕݔ‬ൌ 1. Suppose the new ‫′ݕ′ݔ‬system is formed by rotating ‫-ݔ‬axis and ‫-ݕ‬axis by 45° in anti-clockwise direction.
(a) Find the equation of this conic section in ‫-′ݕ′ݔ‬system.
(b) Hence, identify the surface and sketch the graph.

☺Solution:
(a) Using the transformation formulae:

‫ ݔ‬ൌ ‫ ݔ‬cos 45° െ ‫ ݕ‬sin 45° ൌ




‫ ݕ‬ൌ ‫ ݕ‬ᇱ cos 45° ൅ ‫ ݔ‬ᇱ sin 45° ൌ

1

√2
1

ሺ‫ ݔ‬ᇱ െ ‫ ݕ‬ᇱ ሻ,

√2

ሺ‫ ݔ‬ᇱ ൅ ‫ ݕ‬ᇱ ሻ.

Substitute these into ‫ ݕݔ‬ൌ 1, we get
1 ᇱ
1 ᇱ
1 ᇱଶ
ᇱሻ
ᇱሻ
ሺ‫ ݔ‬െ ‫ݕ‬
ሺ‫ ݔ‬൅ ‫ ݕ‬ൌ 1 ⇒ ൫‫ ݔ‬െ ‫ ݕ‬ᇱ ଶ ൯ ൌ 1
‫ ݕݔ‬ൌ 1 ⇒
2
√2
√2

ψύ

MA1200 Basic Calculus and Linear Algebra I
Lecture Note 1: Coordinate Geometry and Conic Sections

‫ ݔ‬ᇱଶ ‫ ݕ‬ᇱ
‫ ݔ‬ᇱଶ
‫ݕ‬ᇱ


ൌ 1 ሺ‫ ݎ݋‬
ଶെ
ଶ ൌ 1ሻ
2
2
൫√2൯
൫√2൯




(b) From (a). we see that the conic section is a hyperbola (see P.34). The graph (blue line) of this conic section is shown below:
‫ݕ‬

ܸଶ ൌ ሺ√2, 0ሻ
0

ܸଵ ൌ ሺെ√2, 0ሻ

ωτ

45°

MA1200 Basic Calculus and Linear Algebra I
Lecture Note 1: Coordinate Geometry and Conic Sections

‫ݔ‬

Example 14
It is given that the equation of a conic section is

5‫ ݔ‬ଶ ൅ 6‫ ݕݔ‬൅ 5‫ ݕ‬ଶ െ 18√2‫ ݔ‬െ 14√2‫ ݕ‬൅ 26 ൌ 0.

Suppose that the new ‫-′ݕ′ݔ‬system is formed by rotating ‫-ݔ‬axis and ‫-ݕ‬axis by
45° in clockwise direction, find
(a) The equation of the conic section in ‫-′ݕ′ݔ‬coordinates.
(b) Hence, identify the surface and sketch the graph

☺Solution
(a) Using the transformation formula
‫ ݔ‬ൌ ‫ ݔ‬cosሺെ45°ሻ െ ‫ ݕ‬sinሺെ45°ሻ ൌ




‫ ݕ‬ൌ ‫ ݔ‬sinሺെ45°ሻ ൅ ‫ ݕ‬cosሺെ45°ሻ ൌ


ωυ



1

√2

1

√2

ሺ‫ ݔ‬ᇱ ൅ ‫ ݕ‬ᇱ ሻ,

ሺെ‫ ݔ‬ᇱ ൅ ‫ ݕ‬ᇱ ሻ.

MA1200 Basic Calculus and Linear Algebra I
Lecture Note 1: Coordinate Geometry and Conic Sections

Substitute the formulae into the equation, we have

5 ᇱ
6 ᇱ
5
18√2 ᇱ
ᇱ ሻଶ




ᇱ ሻଶ
ሺ‫ ݔ‬൅ ‫ ݕ‬൅ ሺ‫ ݔ‬൅ ‫ ݕ‬ሻሺെ‫ ݔ‬൅ ‫ ݕ‬ሻ ൅ ሺെ‫ ݔ‬൅ ‫ ݕ‬െ
ሺ‫ ݔ‬൅ ‫ ݕ‬ᇱ ሻ
2
2
2
√2
14√2
ሺെ‫ ݔ‬ᇱ ൅ ‫ ݕ‬ᇱ ሻ ൅ 26 ൌ 0

√2
5 ᇱଶ
5 ᇱଶ
ᇱ ᇱ
ᇱଶ
ᇱଶ
ᇱଶ
⇒ ൫‫ ݔ‬൅ 2‫ ݕ ݔ‬൅ ‫ ݕ‬൯ ൅ 3൫െ‫ ݔ‬൅ ‫ ݕ‬൯ ൅ ൫‫ ݔ‬െ 2‫ ݔ‬ᇱ ‫ ݕ‬ᇱ ൅ ‫ ݕ‬ᇱଶ ൯
2
2
െ 18ሺ‫ ݔ‬ᇱ ൅ ‫ ݕ‬ᇱ ሻ െ 14ሺെ‫ ݔ‬ᇱ ൅ ‫ ݕ‬ᇱ ሻ ൅ 26 ൌ 0
⇒ 2‫ ݔ‬ᇱ ൅ 8‫ ݕ‬ᇱ െ 4‫ ݔ‬ᇱ െ 32‫ ݕ‬ᇱ ൅ 26 ൌ 0




⇒ ‫ ݔ‬ᇱ ൅ 4‫ ݕ‬ᇱ െ 2‫ ݔ‬ᇱ െ 16‫ ݕ‬ᇱ ൅ 13 ൌ 0








⇒ ሺ‫ ݔ‬ᇱ െ 1ሻଶ ൅ 4ሺ‫ ݕ‬ᇱ െ 2ሻଶ ൌ 4

ሺ‫ ݔ‬ᇱ െ 1ሻଶ ሺ‫ ݕ‬ᇱ െ 2ሻଶ


ൌ 1.


2
1

ωφ

MA1200 Basic Calculus and Linear Algebra I
Lecture Note 1: Coordinate Geometry and Conic Sections

(b) From the result of (a), we observe that the graph is a standard ellipse (with centre ሺ1,2ሻ in ‫-′ݕ′ݔ‬plane. The graph of the conic section is presented below:
‫ݕ‬

ሺെ1,2ሻ
45°

ሺ1,1ሻ

ሺ1,3ሻ

ሺ3,2ሻ

‫ݔ‬

(Here, coordinates highlighted in red are the coordinates in ‫-′ݕ′ݔ‬plane!!)

ωχ

MA1200 Basic Calculus and Linear Algebra I
Lecture Note 1: Coordinate Geometry and Conic Sections

Identifying the conic section

Given the equation of conic section ‫ ݔܣ‬ଶ ൅ ‫ ݕݔܤ‬൅ ‫ ݕܥ‬ଶ ൅ ‫ ݔܦ‬൅ ‫ ݕܧ‬൅ ‫ ܨ‬ൌ 0, one can identify the conic section using the following procedure:
1. Use the transformation formula
‫ ݔ‬ൌ ‫ ݔ‬ᇱ cos ߠ െ ‫ ݕ‬ᇱ sin ߠ
‫ ݕ‬ൌ ‫ ݔ‬ᇱ sin ߠ ൅ ‫ ݕ‬ᇱ cos ߠ and rewrite the equation into the form

‫ܣ‬ᇱ ‫ ݔ‬ᇱଶ ൅ ‫ܤ‬ᇱ ‫ ݔ‬ᇱ ‫ ݕ‬ᇱ ൅ ‫ ܥ‬ᇱ ‫ ݕ‬ᇱ ൅ ‫ ݔܦ‬ᇱ ൅ ‫ ݕܧ‬ᇱ ൅ ‫ ܨ‬ᇱ ൌ 0. where ‫ܣ‬ᇱ , ‫ܤ‬ᇱ , ‫ ܥ‬ᇱ , ‫ ܦ‬ᇱ , ‫ ܧ‬ᇱ , ‫ ܨ‬ᇱ are some numbers in terms of ߠ.


2. Choose ߠ such that ‫ܤ‬ᇱ ‫ ݔ‬ᇱ ‫ ݕ‬ᇱ ൌ 0 or ‫ܤ‬ᇱ ൌ 0.

3. The equation becomes ‫ܣ‬ᇱ ‫ ݔ‬ᇱ ൅ ‫ ܥ‬ᇱ ‫ ݕ‬ᇱ ൅ ‫ ݔܦ‬ᇱ ൅ ‫ ܧ‬ᇱ ‫ ݕ‬ᇱ ൅ ‫ ܨ‬ᇱ ൌ 0. One can identify the conic section by using the result in P.39.


ωψ



MA1200 Basic Calculus and Linear Algebra I
Lecture Note 1: Coordinate Geometry and Conic Sections

ୀ଴

‫ܤ‬ᇱଶ

Using this procedure and the fact that
െ 4‫ܣ‬ᇱ ‫ ܥ‬ᇱ ൌ ‫ܤ‬ଶ െ 4‫ ,ܥܣ‬one can establish the following characterization of conic section:
Equation of Conic Section

‫ ݔܣ‬ଶ ൅ ‫ ݕݔܤ‬൅ ‫ ݕܥ‬ଶ ൅ ‫ ݔܦ‬൅ ‫ ݕܧ‬൅ ‫ ܨ‬ൌ 0

‫ܤ‬ଶ െ 4‫ ܥܣ‬ൌ 0

Parabola

‫ܤ‬ଶ െ 4‫ ܥܣ‬൐ 0

‫ܤ‬ଶ െ 4‫ ܥܣ‬൏ 0

Hyperbola

Circle/ Ellipse
‫ ܣ‬ൌ ‫ ܥ‬and
‫ܤ‬ൌ0
Circle

ωω

MA1200 Basic Calculus and Linear Algebra I
Lecture Note 1: Coordinate Geometry and Conic Sections

Otherwise

Ellipse

Example 15
Identify the graph of each of the following equations:
(a) ‫ ݔ‬ଶ െ 2‫ ݕݔ‬െ 3‫ ݕ‬ଶ െ 3‫ ݔ‬൅ 6‫ ݕ‬െ 5 ൌ 0
(b) 4‫ ݔ‬ଶ െ 4‫ ݕݔ‬൅ ‫ ݕ‬൅ 2 ൌ 0
(c) 3‫ ݔ‬ଶ െ ‫ ݕݔ‬൅ 12‫ ݕ‬ଶ െ 5‫ ݔ‬െ 7‫ ݕ‬൅ 12 ൌ 0.

☺Solution:

(a) Since ‫ܤ‬ଶ െ 4‫ ܥܣ‬ൌ ሺെ2ሻଶ െ 4ሺ1ሻሺെ3ሻ ൌ 16 ൐ 0, so the graph is hyperbola. (b) Since ‫ܤ‬ଶ െ 4‫ ܥܣ‬ൌ ሺെ4ሻଶ െ 4ሺ4ሻሺ1ሻ ൌ 0, so the graph is parabola.
(c) Since ‫ܤ‬ଶ െ 4‫ ܥܣ‬ൌ ሺെ1ሻଶ െ 4ሺ3ሻሺ12ሻ ൌ െ143 ൏ 0, so the graph can be either ellipse or circle.
Furthermore, since ‫ ܤ‬ൌ െ1 ് 0, we conclude the graph is ellipse.

ωϊ

MA1200 Basic Calculus and Linear Algebra I
Lecture Note 1: Coordinate Geometry and Conic Sections

Parametric Equations
In Physics, we usually describe the motion of a moving particle by the following pair of equations:
‫ ݔ‬ൌ ݂ሺ‫ݐ‬ሻ

, ‫ ∈ ݐ‬ሾܽ, ܾሿ
‫ ݕ‬ൌ ݃ሺ‫ݐ‬ሻ

where ‫ ݐ‬is a parameter (which may be interpreted as time) and ݂ሺ‫ݐ‬ሻ, ݃ሺ‫ݐ‬ሻ are some functions of ‫ .ݐ‬This pair of equation is called parametric equations.
ሺ‫ݕ ,ݔ‬ሻ ൌ ሺ݂ሺ‫ݐ‬ሻ, ݃ሺ‫ݐ‬ሻሻ

ሺ݂ ሺܽሻ, ݃ሺܽሻሻ

ωϋ

MA1200 Basic Calculus and Linear Algebra I
Lecture Note 1: Coordinate Geometry and Conic Sections

ሺ݂ ሺܾሻ, ݃ሺܾሻሻ

Summary – Things you need to know
• Two basic coordinate systems – Cartesian coordinate (‫-ݕݔ‬coordinate) and
Polar coordinate (‫-ߠݎ‬coordinate)
Interchanging between Cartesian coordinate and Polar coordinate.
• Four conic sections – Circle, Ellipse, Parabola, Hyperbola
Know the standard equations of four conic sections
Know how to identify the conic section using completing square technique. • Classification of general conic sections
Classification of ‫ ݔܣ‬ଶ ൅ ‫ ݕܥ‬ଶ ൅ ‫ ݔܦ‬൅ ݁‫ ݕ‬൅ ‫ ܨ‬ൌ 0
Classification of ‫ ݔܣ‬ଶ ൅ ‫ ݕݔܤ‬൅ ‫ ݕܥ‬ଶ ൅ ‫ ݔܦ‬൅ ‫ ݕܧ‬൅ ‫ ܨ‬ൌ 0
- Using rotation of axes
- Using ‫ܤ‬ଶ െ 4‫.ܥܣ‬

ωό

MA1200 Basic Calculus and Linear Algebra I
Lecture Note 1: Coordinate Geometry and Conic Sections

Appendix – Detailed description of the classification procedure of ࡭࢞૛ ൅
࡮࢞࢟ ൅ ࡯࢟૛ ൅ ࡰ࢞ ൅ ࡱ࢟ ൅ ࡲ ൌ ૙. (For ࡮ ് ૙)

This appendix provides a detailed explanation of the classification procedure described in P.54.
Step 1: Transform the equation

Apply the transformation formula

‫ ݔ‬ൌ ‫ ݔ‬ᇱ cos ߠ െ ‫ ݕ‬ᇱ sin ߠ ܽ݊݀ ‫ ݕ‬ൌ ‫ ݔ‬ᇱ sin ߠ ൅ ‫ ݕ‬ᇱ cos ߠ,

one can transform the equation into the alternative form

‫ܣ‬ᇱ ‫ ݔ‬ᇱ ൅ ‫ܤ‬ᇱ ‫ ݔ‬ᇱ ‫ ݕ‬ᇱ ൅ ‫ ܥ‬ᇱ ‫ ݕ‬ᇱଶ ൅ ‫ ܦ‬ᇱ ‫ ݔ‬ᇱ ൅ ‫ ܧ‬ᇱ ‫ ݕ‬ᇱ ൅ ‫ ܨ‬ᇱ ൌ 0


where

‫ܣ‬ᇱ ൌ ‫ ܣ‬cos ଶ ߠ ൅ ‫ ܤ‬cos ߠ sin ߠ ൅ ‫ ܥ‬sinଶ ߠ,

‫ܤ‬ᇱ ൌ െ2‫ ܣ‬cos ߠ sin ߠ ൅ ‫ܤ‬ሺcos ଶ ߠ െ sinଶ ߠሻ ൅ 2‫ ܥ‬sin ߠ cos ߠ

ωύ

MA1200 Basic Calculus and Linear Algebra I
Lecture Note 1: Coordinate Geometry and Conic Sections

‫ ܥ‬ᇱ ൌ ‫ ܣ‬sinଶ ߠ െ ‫ ܤ‬sin ߠ cos ߠ ൅ ‫ ܥ‬cosଶ ߠ

‫ ܦ‬ᇱ ൌ ‫ ܦ‬cos ߠ ൅ ‫ ܧ‬sin ߠ ,
Step 2:

‫ ܧ‬ᇱ ൌ െ‫ ܦ‬sin ߠ ൅ ‫ ܧ‬cos ߠ ,

‫ ܨ‬ᇱ ൌ ‫.ܨ‬

Choose ߠ such that ‫ܤ‬ᇱ ൌ 0 so that the equation becomes ‫ܣ‬ᇱ ‫ ݔ‬ᇱ ൅ ‫ ܥ‬ᇱ ‫ ݕ‬ᇱ ൅
‫ ܦ‬ᇱ ‫ ݔ‬ᇱ ൅ ‫ ܧ‬ᇱ ‫ ݕ‬ᇱ ൅ ‫ ܨ‬ᇱ ൌ 0 and the result in P.39 can be applied.


We need to solve



‫ܤ‬ᇱ ൌ െ2‫ ܣ‬cos ߠ sin ߠ ൅ ‫ܤ‬ሺcosଶ ߠ െ sinଶ ߠሻ ൅ 2‫ ܥ‬sin ߠ cos ߠ ൌ 0.

Using the fact that

we get

2 sin ߠ cos ߠ ൌ sin 2ߠ ܽ݊݀ cosଶ ߠ െ sinଶ ߠ ൌ cos 2ߠ,
െ‫ ܣ‬sin 2ߠ ൅ ‫ ܤ‬cos 2ߠ ൅ ‫ ܥ‬sin 2ߠ ൌ 0
⇒ ሺ‫ ܣ‬െ ‫ ܥ‬ሻ sin 2ߠ ൌ ‫ ܤ‬cos 2ߠ

ϊτ

MA1200 Basic Calculus and Linear Algebra I
Lecture Note 1: Coordinate Geometry and Conic Sections

‫ܤ‬ ݂݅ ‫ܥ ് ܣ‬
⇒൝
‫ܣ‬െ‫ܥ‬ cos 2ߠ ൌ 0 ݂݅ ‫ ܣ‬ൌ ‫ ܥ‬ tan 2ߠ ൌ

With this ߠ, the equation becomes

‫ܣ‬ᇱ ‫ ݔ‬ᇱ ൅ ‫ ܥ‬ᇱ ‫ ݕ‬ᇱ ൅ ‫ ܦ‬ᇱ ‫ ݔ‬ᇱ ൅ ‫ ܧ‬ᇱ ‫ ݕ‬ᇱ ൅ ‫ ܨ‬ᇱ ൌ 0.




The result in P.39 suggests that

‫ܣ‬ᇱ ‫ ܥ‬ᇱ ൌ 0 ⇒ The conic section is parabola

‫ܣ‬ᇱ ‫ ܥ‬ᇱ ൐ 0 ⇒ The conic section is ellipse or circle
‫ܣ‬ᇱ ‫ ܥ‬ᇱ ൏ 0 ⇒ The conic section is hyperbola

(* Here, we do not consider those “degenerate” cases)

ϊυ

MA1200 Basic Calculus and Linear Algebra I
Lecture Note 1: Coordinate Geometry and Conic Sections

Step 4: Establish the relationship between ‫ ܥ ,ܤ ,ܣ‬and ‫ܣ‬ᇱ , ‫ܤ‬ᇱ , ‫ ܥ‬ᇱ

One can show, by direct (but tedious) calculation, that

‫ܤ‬ᇱ െ 4‫ܣ‬ᇱ ‫ ܥ‬ᇱ ൌ ‫ܤ‬ଶ െ 4‫.ߠ ݕ݊ܽ ݎ݋݂ ܥܣ‬


Since the ߠ is chosen such that ‫ܤ‬ᇱ ൌ 0, so we have

‫ܤ‬ଶ െ 4‫ ܥܣ‬ൌ െ4‫ܣ‬ᇱ ‫ ܥ‬ᇱ .

Therefore, we can conclude that

‫ܤ‬ଶ െ 4‫ ܥܣ‬ൌ 0 ⇒ െ4‫ܣ‬ᇱ ‫ ܥ‬ᇱ ൌ 0 ⇒ ‫ܣ‬ᇱ ‫ ܥ‬ᇱ ൌ 0 ⇒ parabola

‫ܤ‬ଶ െ 4‫ ܥܣ‬൐ 0 ⇒ െ4‫ܣ‬ᇱ ‫ ܥ‬ᇱ ൐ 0 ⇒ ‫ܣ‬ᇱ ‫ ܥ‬ᇱ ൏ 0 ⇒ hyperbola

‫ܤ‬ଶ െ 4‫ ܥܣ‬൏ 0 ⇒ െ4‫ܣ‬ᇱ ‫ ܥ‬ᇱ ൏ 0 ⇒ ‫ܣ‬ᇱ ‫ ܥ‬ᇱ ൐ 0 ⇒ circle or ellipse.

(*In the last case, one can check that the conic is circle only when ‫ ܤ‬ൌ 0 and
‫ ܣ‬ൌ ‫)ܥ‬

ϊφ

MA1200 Basic Calculus and Linear Algebra I
Lecture Note 1: Coordinate Geometry and Conic Sections

One more examples on hyperbola and parabola
Extra Example 1 (Parabola: Degenerate Case)

It is given that the equation of a parabola is ‫ ݕ‬ଶ ൅ ܽ‫ ݔ‬െ 4‫ ݕ‬൅ 3 ൌ 0.
(a) If ܽ ൌ 1, sketch the graph and locate the vertex of parabola.
(b) What will happen when ܽ ൌ 0?

☺Solution:
(a) If ܽ ൌ 1, the equation becomes
‫ ݕ‬ଶ ൅ ‫ ݔ‬െ 4‫ ݕ‬൅ 3 ൌ 0
⇒ ሺ‫ ݕ‬ଶ െ 4‫ݕ‬ሻ ൅ ‫ ݔ‬൅ 3 ൌ 0
⇒ ሺ‫ ݕ‬ଶ െ 2ሺ2ሻ‫ ݕ‬൅ 2ଶ െ 2ଶ ሻ ൅ ‫ ݔ‬൅ 3 ൌ 0
⇒ ሺ‫ ݕ‬െ 2ሻଶ ൅ ‫ ݔ‬െ 1 ൌ 0 ⇒ ሺ‫ ݕ‬െ 2ሻଶ ൌ െሺ‫ ݔ‬െ 1ሻ
1

ሺ‫ ݕ‬െ 2ሻ
⇒ ᇣᇧᇤᇧᇥ ൌ 4 ൬െ ൰ ሺ‫ ݔ‬ᇤ 1ሻ
ᇣ െᇧ
ᇧ ᇥ
ᇣᇤᇥ
4 ሺ௫ି௛ሻ

ሺ௬ି௞ ሻ



This parabola has vertex ሺ݄, ݇ሻ ൌ ሺ1,2ሻ.

ϊχ

MA1200 Basic Calculus and Linear Algebra I
Lecture Note 1: Coordinate Geometry and Conic Sections

(b) If ܽ ൌ 0, the equation becomes
‫ ݕ‬ଶ െ 4‫ ݕ‬൅ 3 ൌ 0.
One can solve this equation and get
‫ ݕ‬ଶ െ 4‫ ݕ‬൅ 3 ൌ 0 ⇒ ሺ‫ ݕ‬െ 1ሻሺ‫ ݕ‬െ 3ሻ ൌ 0 ⇒ ‫ ݕ‬ൌ 1 ‫ ݕ ݎ݋‬ൌ 3.
Then the conic section becomes two separated horizontal lines on 2-D plane. ☺Note:
The above case is called “degenerate case” .

ϊψ

MA1200 Basic Calculus and Linear Algebra I
Lecture Note 1: Coordinate Geometry and Conic Sections

Extra Example 2

It is given that the equation of a conic section is 3‫ ݔ‬ଶ ൅ 2√3‫ ݕݔ‬൅ ‫ ݕ‬ଶ െ ‫ ݔ‬൅
√3‫ ݕ‬൅ 4 ൌ 0.
(a) Identify the conic section (Circle/ Ellipse/ Hyperbola or Parabola)
(b) Sketch the graph and indicate the necessary details.

☺Solution
(a) Note that



‫ܤ‬ଶ െ 4‫ ܥܣ‬ൌ ൫2√3൯ െ 4ሺ3ሻሺ1ሻ ൌ 12 െ 12 ൌ 0.
So the conic section is parabola.

(b) We need to transform the equation into the form: ‫ܣ‬ᇱ ‫ ݔ‬ᇱ ൅ ‫ ܥ‬ᇱ ‫ ݕ‬ᇱ ൅
‫ ′ݔ′ܦ‬൅ ‫ ′ݕ′ܧ‬൅ ‫ ′ܨ‬ൌ 0 (using rotation of axes).


Since tan 2ߠ ൌ

ϊω



஺ି஼

⇒ tan 2ߠ ൌ

ଶ√ଷ
ଷିଵ



ൌ √3 ⇒ 2ߠ ൌ 60° ⇒ ߠ ൌ 30°.

MA1200 Basic Calculus and Linear Algebra I
Lecture Note 1: Coordinate Geometry and Conic Sections

Therefore one can transform the equation into the desired form by rotating ‫-ݕݔ‬axes by 30° in anti-clockwise direction.
√3 ᇱ 1 ᇱ
‫ ݔ‬ൌ ‫ ݔ‬cosሺ30°ሻ െ ‫ ݕ‬sinሺ30°ሻ ൌ
‫ ݔ‬െ ‫ݕ‬
2
2
1
√3 ᇱ
‫ ݕ‬ൌ ‫ ݔ‬ᇱ sinሺ30°ሻ ൅ ‫ ݕ‬ᇱ cosሺ30°ሻ ൌ ‫ ݔ‬ᇱ ൅
‫,ݕ‬
2
2
and substitute into the original equation, we get

Using the transformation formula:





ଵ ᇱ


√ଷ ᇱ
√ଷ
√ଷ
3 ቀ ‫ ݔ‬െ ‫ ݕ‬ቁ ൅ 2√3 ቀ ‫ ݔ‬ᇱ െ ‫ ݕ‬ᇱ ቁ ቀ ‫ ݔ‬ᇱ ൅ ‫ ݕ‬ᇱ ቁ









√ଷ ᇱ
√ଷ
√ଷ
‫ ݕ‬ቁ െ ቀ ‫ ݔ‬ᇱ െ ‫ ݕ‬ᇱ ቁ ൅ √3 ቀ ‫ ݔ‬ᇱ ൅ ‫ ݕ‬ᇱ ቁ ൅ 4 ൌ 0






⇒⋯

⇒ 4‫ ݔ‬ᇱ ൅ 2‫ ݕ‬ᇱ ൅ 4 ൌ 0

ϊϊ



MA1200 Basic Calculus and Linear Algebra I
Lecture Note 1: Coordinate Geometry and Conic Sections

൅ ቀ ‫ݔ‬ᇱ ൅



1
⇒ ‫ ݔ‬ᇱଶ ൌ െ ሺ2‫ ݕ‬ᇱ ൅ 4ሻ
4
1 ᇱ
ᇱଶ
⇒ ‫ ݔ‬ൌ െ ሺ‫ ݕ‬൅ 2ሻ
2
1
ᇱଶ
⇒ ‫ݔ‬
ด ൌ െ4 ൬ ൰ ൫‫ ݕ‬ᇱ ᇧᇤᇧᇧᇥ
ᇣᇧ െ ሺെ2ሻ൯
ต ᇧ ᇧ
8
ᇲ ି௛ሻమ
ሺ௫



ሺ௬ ି௞ሻ

So the vertex of the parabola is ሺ݄, ݇ሻ ൌ ሺ0, െ2ሻ and it is the inverted Ushape curve in ‫ ݔ‬ᇱ ‫ ݕ‬ᇱ -plane.

ϊϋ

MA1200 Basic Calculus and Linear Algebra I
Lecture Note 1: Coordinate Geometry and Conic Sections

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...Math 1P05 Assignment #1 Due: September 26 Questions 3, 4, 6, 7, 11 and 12 require some Maple work. 1. Solve the following inequalities: a) b) c) 2. Appendix D #72 3. Consider the functions and . a) Use a Maple graph to estimate the largest value of at which the graphs intersect. Hand in a graph that clearly shows this intersection. b) Use Maple to help you find all solutions of the equation. 4. Consider the function. a) Find the domain of. b) Find and its domain. What is the range of? c) To check your result in b), plot and the line on the same set of axes. (Hint: To get a nice graph, choose a plotting range for bothand.) Be sure to label each curve. 5. Section 1.6 #62 6. Section 2.1 #4. In d), use Maple to plot the curve and the tangent line. Draw the secant lines by hand on your Maple graph. 7. Section 2.2 #24. Use Maple to plot the function. 8. Section 2.2 #36 9. Section 2.3 #14 10. Section 2.3 #26 11. Section 2.3 #34 12. Section 2.3 #36 Recommended Problems Appendix A all odd-numbered exercises 1-37, 47-55 Appendix B all odd-numbered exercises 21-35 Appendix D all odd-numbered exercises 23-33, 65-71 Section 1.5 #19, 21 Section 1.6 all odd-numbered exercises 15-25, 35-41, 51, 53 Section 2.1 #3, 5, 7 Section 2.2 all odd-numbered exercises 5-9, 15-25, 29-37 Section 2.3 all odd-numbered exercises...

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...find the national average cost of food for an individual, as well as for a family of 4 for a given month. http://www.cnpp.usda.gov/sites/default/files/usda_food_plans_cost_of_food/CostofFoodJan2012.pdf 5. Find a website for your local city government. http://www.usa.gov/Agencies/Local.shtml 6. Find the website for your favorite sports team (state what that team is as well by the link). http://blackhawks.nhl.com/ (Chicago Blackhawks) 7. Many of us do not realize how often we use math in our daily lives. Many of us believe that math is learned in classes, and often forgotten, as we do not practice it in the real world. Truth is, we actually use math every day, all of the time. Math is used everywhere, in each of our lives. Math does not always need to be thought of as rocket science. Math is such a large part of our lives, we do not even notice we are computing problems in our lives! For example, if one were interested in baking, one must understand that math is involved. One may ask, “How is math involved with cooking?” Fractions are needed to bake an item. A real world problem for baking could be as such: Heena is baking a cake that requires two and one-half cups of flour. Heena poured four and one-sixth cups of flour into a bowl. How much flour should Heena take out of the bowl? In this scenario of a real world problem, we have fractions, and subtraction of fractions, since Heena has added four and one-sixth cups of flour, rather than the needed...

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... h(x)= 7-x/3 First we need to compute (f-h)(4) (f*h)(4)=f(4)-h(4), each function can be done separately f(4)=2(4)+5 f(4)=8+5 f(4)=13 H h(4)=(7-4)/3 same process as above h(4)=3/3=h(4)=1 (f-h)(4)=13-1 (f-h)(4)=12 this is the solution after substituting and subtracting The next part we need to replace the x in the f function with the g (f*g)(x)=f(g(x)) (f*g)(x)=f(x2-3) (f*g)(x)=2x2-1 is the result Now we need to do the h function (h*g)(x)=h(g(x)) (h*g)(x)=h(x2-3) (h*g)(x)=7-(x2-3) (h*g)(x)=10-x2 end result The inverse function-- f-1(x)=x-5h-1(x)=-(3-7) By doing problems this way it can save a person and a business a lot of time. A lot of people think they don't need math everyday throughout their life, but in all reality people use math almost everyday in life. The more you know the better off your life will...

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