Preliminary Mathematics (2U)

Preliminary Mathematics (2U)

Locus – Parabolas

Week 2

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Preliminary Mathematics (2U) 2

Locus – Parabolas

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Week 2 – Theory A Parabola:

A parabola may be defined as the locus of a point 𝑃(𝑥, 𝑦) whose distance from a given fixed point

equals its distance from a given fixed line. The fixed point is known as the focus and the fixed line is

known as the directrix.

The vertex is the minimum or maximum point of the parabola. The axis of symmetry is a line which

bisects the parabola. The focal length is the distance between the vertex and the focus or the vertex

and the directrix.


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The General Equation of the Locus of a Parabola with Vertex at the Origin:

Case One (Example):

Find the locus of a point 𝑃(𝑥, 𝑦) such that its distance from the point 𝐹(0, 𝑎) is equal to its distance

from the line 𝑦 = −𝑎.

Solution:

Given the distance of 𝐹𝑃 = 𝐴𝑃, so use the distance formula:

𝐹𝑃 = 𝐴𝑃

√(𝑥 − 0)2 + (𝑦 − 𝑎)2 = √(𝑥 − 𝑥)2 + (𝑦 − −𝑎)2

𝑥2 + (𝑦 − 𝑎)2 = 02 + (𝑦 + 𝑎)2

𝑥2 + 𝑦2 − 2𝑎𝑦 + 𝑎2 = 𝑦2 + 2𝑎𝑦 + 𝑎2

∴ 𝑥2 = 4𝑎𝑦


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Case Two (Example):

Find the locus of a point 𝑃(𝑥, 𝑦) such that its distance from the point 𝐹(0, −𝑎) is equal to its

distance from the line 𝑦 = 𝑎.

Solution:


𝐹𝑃 = 𝐴𝑃

√(𝑥 − 0)2 + (𝑦 − −𝑎)2 = √(𝑥 − 𝑥)2 + (𝑦 − 𝑎)2

𝑥2 + (𝑦 + 𝑎)2 = 02 + (𝑦 − 𝑎)2

𝑥2 + 𝑦2 + 2𝑎𝑦 + 𝑎2 = 𝑦2 − 2𝑎𝑦 + 𝑎2

∴ 𝑥2 = −4𝑎𝑦


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Case Three (Example):

Find the locus of a point 𝑃(𝑥, 𝑦) such that its distance from the point 𝐹(𝑎, 0) is equal to its distance

from the line 𝑥 = −𝑎.

Solution:


𝐹𝑃 = 𝐴𝑃

√(𝑥 − 𝑎)2 + (𝑦 − 0)2 = √(𝑥 − −𝑎)2 + (𝑦 − 𝑦)2

(𝑥 − 𝑎)2 + 𝑦2 = (𝑥 + 𝑎)2 + 02

𝑥2 − 2𝑎𝑥 + 𝑎2 + 𝑦2 = 𝑥2 + 2𝑎𝑥 + 𝑎2

∴ 𝑦2 = 4𝑎𝑥


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Case Four (Example):

Find the locus of a point 𝑃(𝑥, 𝑦) such that its distance from the point 𝐹(−𝑎, 0) is equal to its

distance from the line 𝑥 = 𝑎.

Solution:


𝐹𝑃 = 𝐴𝑃

√(𝑥 − −𝑎)2 + (𝑦 − 0)2 = √(𝑥 − 𝑎)2 + (𝑦 − 𝑦)2

(𝑥 + 𝑎)2 + 𝑦2 = (𝑥 − 𝑎)2 + 02

𝑥2 + 2𝑎𝑥 + 𝑎2 + 𝑦2 = 𝑥2 − 2𝑎𝑥 + 𝑎2

∴ 𝑦2 = −4𝑎𝑥


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Example:

Sketch the parabola with equation 𝑥2 = −16𝑦 and then find:

(i) The focal length,

(ii) The coordinates of the vertex,

(iii) The coordinates of the focus,

(iv) The equation of the directrix,

(v) The equation of the axis of symmetry.

Solution:

𝑥2 = −16𝑦

(i) The focal length:

4𝑎 = 16

∴ 𝑎 = 4

∴ The focal length is 4 units.

(ii) The coordinates of the vertex is (0, 0).

(iii) The coordinates of the focus is (0, −4).

(iv) The equation of the directrix is 𝑦 = 4.

(v) The equation of the axis of symmetry is 𝑥 = 0.


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The General Equation of the Locus of a Parabola with Vertex not at the Origin:

Case One (Example):

Find the locus of a point 𝑃(𝑥, 𝑦) such that its distance from the point 𝐹(ℎ, 𝑘 + 𝑎) is equal to its

distance from the line 𝑦 = 𝑘 − 𝑎.

Solution:


𝐹𝑃 = 𝐴𝑃

√(𝑥 − ℎ)2 + (𝑦 − (𝑘 + 𝑎))2

= √(𝑥 − 𝑥)2 + (𝑦 − (𝑘 − 𝑎))2

(𝑥 − ℎ)2 + (𝑦 − (𝑘 + 𝑎))2

= (𝑦 − (𝑘 − 𝑎))2

(𝑥 − ℎ)2 + 𝑦2 − 2(𝑘 + 𝑎)𝑦 + (𝑘 + 𝑎)2 = 𝑦2 − 2(𝑘 − 𝑎)𝑦 + (𝑘 − 𝑎)2

(𝑥 − ℎ)2 + 𝑦2 − 2𝑘𝑦 − 2𝑎𝑦 + 𝑘2 + 2𝑎𝑘 + 𝑎2 = 𝑦2 − 2𝑘𝑦 + 2𝑎𝑦 + 𝑘2 − 2𝑎𝑘 + 𝑎2

(𝑥 − ℎ)2 = 4𝑎𝑦 − 4𝑎𝑘

∴ (𝑥 − ℎ)2 = 4𝑎(𝑦 − 𝑘)


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Case Two (Example):

Find the locus of a point 𝑃(𝑥, 𝑦) such that its distance from the point 𝐹(ℎ, 𝑘 − 𝑎) is equal to its

distance from the line 𝑦 = 𝑘 + 𝑎.

Solution:


𝐹𝑃 = 𝐴𝑃

√(𝑥 − ℎ)2 + (𝑦 − (𝑘 − 𝑎))2

= √(𝑥 − 𝑥)2 + (𝑦 − (𝑘 + 𝑎))2

(𝑥 − ℎ)2 + (𝑦 − (𝑘 − 𝑎))2

= (𝑦 − (𝑘 + 𝑎))2

(𝑥 − ℎ)2 + 𝑦2 − 2(𝑘 − 𝑎)𝑦 + (𝑘 − 𝑎)2 = 𝑦2 − 2(𝑘 + 𝑎)𝑦 + (𝑘 + 𝑎)2

(𝑥 − ℎ)2 + 𝑦2 − 2𝑘𝑦 + 2𝑎𝑦 + 𝑘2 − 2𝑎𝑘 + 𝑎2 = 𝑦2 − 2𝑘𝑦 − 2𝑎𝑦 + 𝑘2 + 2𝑎𝑘 + 𝑎2

(𝑥 − ℎ)2 = −4𝑎𝑦 + 4𝑎𝑘

∴ (𝑥 − ℎ)2 = −4𝑎(𝑦 − 𝑘)


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Case Three (Example):

Find the locus of a point 𝑃(𝑥, 𝑦) such that its distance from the point 𝐹(ℎ + 𝑎, 𝑘) is equal to its

distance from the line 𝑥 = ℎ − 𝑎.

Solution:


𝐹𝑃 = 𝐴𝑃

√(𝑥 − (ℎ + 𝑎))2

+ (𝑦 − 𝑘)2 = √(𝑥 − (ℎ − 𝑎))2

+ (𝑦 − 𝑦)2

(𝑥 − (ℎ + 𝑎))2

+ (𝑦 − 𝑘)2 = (𝑥 − (ℎ − 𝑎))2

𝑥2 − 2(ℎ + 𝑎)𝑥 + (ℎ + 𝑎)2 + (𝑦 − 𝑘)2 = 𝑥2 − 2(ℎ − 𝑎)𝑥 + (ℎ − 𝑎)2

𝑥2 − 2ℎ𝑥 − 2𝑎𝑥 + ℎ2 + 2𝑎ℎ + 𝑎2 + (𝑦 − 𝑘)2 = 𝑥2 − 2ℎ𝑥 + 2𝑎𝑥 + ℎ2 − 2𝑎ℎ + 𝑎2

(𝑦 − 𝑘)2 = 4𝑎𝑥 − 4𝑎ℎ

∴ (𝑦 − 𝑘)2 = 4𝑎(𝑥 − ℎ)


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Case Four (Example):

Find the locus of a point 𝑃(𝑥, 𝑦) such that its distance from the point 𝐹(ℎ − 𝑎, 𝑘) is equal to its

distance from the line 𝑥 = ℎ + 𝑎.

Solution:


𝐹𝑃 = 𝐴𝑃

√(𝑥 − (ℎ − 𝑎))2

+ (𝑦 − 𝑘)2 = √(𝑥 − (ℎ + 𝑎))2

+ (𝑦 − 𝑦)2

(𝑥 − (ℎ − 𝑎))2

+ (𝑦 − 𝑘)2 = (𝑥 − (ℎ + 𝑎))2

𝑥2 − 2(ℎ − 𝑎)𝑥 + (ℎ − 𝑎)2 + (𝑦 − 𝑘)2 = 𝑥2 − 2(ℎ + 𝑎)𝑥 + (ℎ + 𝑎)2

𝑥2 − 2ℎ𝑥 + 2𝑎𝑥 + ℎ2 − 2𝑎ℎ + 𝑎2 + (𝑦 − 𝑘)2 = 𝑥2 − 2ℎ𝑥 − 2𝑎𝑥 + ℎ2 + 2𝑎ℎ + 𝑎2

(𝑦 − 𝑘)2 = −4𝑎𝑥 + 4𝑎ℎ

∴ (𝑦 − 𝑘)2 = −4𝑎(𝑥 − ℎ)


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Example:

Find the equation of the locus of a point 𝑃(𝑥, 𝑦) such that its distance from the point 𝐹(3, 2) is equal

to its distance from the line 𝑦 = −6.

Solution:

Firstly, sketch a diagram.


𝐹𝑃 = 𝐴𝑃

√(𝑥 − 3)2 + (𝑦 − 2)2 = √(𝑥 − 𝑥)2 + (𝑦 + 6)2

(𝑥 − 3)2 + (𝑦 − 2)2 = (𝑦 + 6)2

(𝑥 − 3)2 + 𝑦2 − 4𝑦 + 4 = 𝑦2 + 12𝑦 + 36

(𝑥 − 3)2 = 16𝑦 + 32

∴ (𝑥 − 3)2 = 16(𝑦 + 2)


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Example:

For the parabola with equation 2𝑦2 − 8𝑦 + 𝑥 + 9 = 0, find:

(i) The coordinates of the vertex and focus.

(ii) The equation of the directrix and the axis of symmetry.

(iii) Hence, sketch the parabola.

Solution:

(i)

2𝑦2 − 8𝑦 + 𝑥 + 9 = 0

𝑦2 − 4𝑦 +1

2𝑥 +

9

2= 0

𝑦2 − 4𝑦 + (−2)2 = −1

2𝑥 −

9

2+ (−2)2

(𝑦 − 2)2 = −1

2𝑥 −

1

2

(𝑦 − 2)2 = −1

2(𝑥 + 1)

The focal length:

4𝑎 =1

2

𝑎 =1

8

∴ The vertex is (−1, 2) and the focus is (−11

8, 2).

(ii) The equation of the directrix is 𝑥 = −7

8 and the axis of symmetry is 𝑦 = 2.

(iii)


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Example:

Find the equation of the parabola with vertex (2, 3) and the equation of the directrix is 𝑥 = 3.

Solution:

Firstly, sketch a diagram.

Hence, the focal length is 1.

(𝑦 − 𝑘)2 = −4𝑎(𝑥 − ℎ)

(𝑦 − 3)2 = −4(1)(𝑥 − 2)

∴ (𝑦 − 3)2 = −4(𝑥 − 2)


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Week 2 – Homework Locus of a Parabola:

1. Sketch the following parabolas, then find:

(i) The coordinates of the vertex,

(ii) The coordinates of the focus,

(iii) The equation of the directrix,

(iv) The equation of the axis of symmetry.

a) 𝑥2 = 4𝑦 b) 𝑥2 = 𝑦 c) 𝑥2 = −0.5𝑦 d) 𝑦2 = −4𝑥

e) 𝑦2 = −1

6𝑥

f) (𝑥 + 1)2 = 4(𝑦 − 1)

g) (𝑥 − 3)2 = −1

16𝑦

h) (𝑦 − 2)2 = 8(𝑥 + 1)

i) (𝑦 + 3)2 = −1

2(𝑥 − 2)


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2. Find the locus of a point 𝑃(𝑥, 𝑦) such that:

a) Its distance from the point (0, 4) is equal to its distance from the line 𝑦 = −4.

b) Its distance from the point (2, 0) is equal to its distance from the line 𝑥 = −2.

c) Its distance from the point (1, −1) is equal to its distance from the line 𝑦 + 3 = 0.

d) Its distance from the point (−7

4, −4) is equal to its distance from the line 𝑥 = −

1

4.


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3. Find:

(i) The coordinates of the vertex and focus.

(ii) The equation of the directrix and the axis of symmetry for the following.

(iii) Hence, sketch the parabolas.

a) 𝑥2 − 2𝑥 − 𝑦 + 4 = 0 b) 𝑦2 − 6𝑦 + 8𝑥 + 1 = 0

c) 2𝑥2 − 8𝑥 + 𝑦 + 6 = 0 d) 4𝑦2 + 24𝑦 − 𝑥 + 38 = 0


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4. Find the equation of the following parabolas with:

a) Vertex at the origin and the focal length is 1 unit.

b) Vertex at the origin, the focal length is 1

2 unit and the axis of symmetry is the 𝑥-axis.

c) Vertex at (1, 2) and the equation of the directrix is 𝑥 = 2.

d) Vertex at (−2, −4) and the focus is (−2, 1).

e) Focus at (0, 0) and the equation of the directrix is 𝑥 = 4.

f) Vertex at (−1, 2), the axis of symmetry is 𝑦 = 2 and the 𝑥-intercept is (−3

2, 0).

End of homework

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