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Year 11 Math Homework

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Table of contents 11 Year 11 Topic 11 — The Geometry of the Parabola (Part1)

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11.1 A Locus and its Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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11.1.1 Finding the locus algebraically . . . . . . . . . . . . . . . . . . . . . . . . .

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11.1.2 Finding the equation of a circle . . . . . . . . . . . . . . . . . . . . . . . .

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11.2 The Geometric Definition of the Parabola . . . . . . . . . . . . . . . . . . . . . . .

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11.2.1 Using the definition of a parabola to find its equation . . . . . . . . . . . . .

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11.2.2 Vertex, Axis of symmetry, Latus rectum and Focal length . . . . . . . . . . .

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11.2.3 Local length, Chords and Focal chords . . . . . . . . . . . . . . . . . . . . .

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11.2.4 The four standard positions of the parabola . . . . . . . . . . . . . . . . . .

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This edition was printed on August 27, 2010. Camera ready copy was prepared with the LATEX2e typesetting system. Copyright © 2000 - 2010 Yimin Math Centre (www.yiminmathcentre.com) Year 11 Math Homework

Year 11 Term 3 Week 1 Homework

11

Page 1 of 5

Year 11 Topic 11 — The Geometry of the Parabola (Part1)

11.1

A Locus and its Equation

11.1.1

Finding the locus algebraically

• Let P (x, y) be any point in the plane. • the point lies on the locus if and only if its coordinates satisfy the equation. • when the distance formula is used, it is best to square the geometric condition first. 11.1.2

Finding the equation of a circle

• A circle is the locus of all points in the plane that are a fixed distance from a given point. • where the fixed distance is called the radius and the given point is called the centre. Example 11.1.1 Find the equation of the locus of a point which moves so that its distance from the point (1, 2) is twice its distance from the point B(-5, -4). Describe the locus geometrically. Solution: Let P (x, y) be any point in the plane. So the condition that P lie on the locus is: P A = 2P B, square it we have P A2 = 4P B 2 (x − 1)2 + (y − 2)2 = 4 × [(x + 5)2 + (y + 4)2 ] x2 − 2x + 1 + y 2 − 4y + 4 = 4(x2 + 10x + 25) + 4(y 2 + 8y + 16) x2 − 2x + 1 + y 2 − 4y + 4 = 4x2 + 40x + 100 + 4y 2 + 32y + 64 3x2 + 42x + y 2 + 36y = −159; then dividing both sides by 3 and completing both squares: x2 + 14x + 49 + y 2 + 12y + 36 = −53 + 49 + 36 √ √ (x+7)2 +(y +6)2 = 32, So the locus is a circle with centre (-7, -6) and radius is 32 or r = 4 2

Exercise 11.1.1 Derive the equation of the locus of the point P (x, y) which moves so that it is always a distance of 4 units from the point A(−2, 1)

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Year 11 Term 3 Week 1 Homework

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Exercise 11.1.2 Use the distance formula to find the equation of the locus of the point P (x, y) which moves so that it is equidistant from the points Q(−2, 4) and R(1, 2).

2. Find the equation of the line through the midpoint of QR and perpendicular to QR. Comment on the result.

Exercise 11.1.3 1. By using the perpendicular distance formula, find the locus of a point P (x, y) which is equidistant from the lines 3x + 4y = 36 and 4x + 3y = 24.

2. Show that the locus consists of two perpendicular lines, and sketch all four lines on the same number plane.

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Year 11 Term 3 Week 1 Homework

11.2

The Geometric Definition of the Parabola

11.2.1

Using the definition of a parabola to find its equation

Page 3 of 5

Example 11.2.1 Find the equation of the locus of all points equidistance from the point S(4, 3) and the line d: y = −3

Solution: Let P(x, y) be any point in the plane, and let M(x, -3) be the foot of the perpendicular from P to d. The condition that P lie on the locus is P S = P M square it both sides: P S 2 = P M 2 (x − 4)2 + (y − 3)2 = (y + 3)2 (x − 4)2 + y 2 − 6y + 9 = y 2 + 6y + 9; (x − 4)2 = 12y

Exercise 11.2.1 The variable point P (x, y) moves so that it is equidistant from point S(0, 3) and the line y + 3 = 0. Draw a diagram and let L be the point (x, −3). 1. Show that P S 2 = x2 + (y − 3)2 and P L2 = (y + 3)2 .

2. By setting P S 2 = P L2 , derive the equation of the locus of P

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Year 11 Term 3 Week 1 Homework 11.2.2

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Vertex, Axis of symmetry, Latus rectum and Focal length

• A parabola is the locus of all points that are equidistant from a given point (called the focus) and given line (called directrix) not passing through the focus. • The vertex V is the point midway between the focus and the directrix. • The line through the vertex V parallel to the directrix is the tangent to the parabola at the vertex. • The line SV through the focus and vertex perpendicular to the directrix is called the axis of symmetry. • The interval joining the points A and B passes through the focus and its parallel to directrix is called the latus rectum.

11.2.3

Local length, Chords and Focal chords

• Let the distance from focus to vertex = a (the focal length). • Then distance from focus to directrix = 2a (twice focal length). • Length of latus rectum = 4a. Example 11.2.2 Use the definition of the parabola to find the equation of the parabola with focus S(0, 2) and directrix d: y = −2. Find the vertex, focal length and length of the latus rectum.

Solution: Let P (x, y) be any point in the plane, and let M (x, −2) be the foot of the perpendicular from P to d. The condition that P lie on the parabola is P S = P M , Square both sides P S 2 = P M 2 x2 + (y − 2)2 = (y + 2)2 ; x2 + y 2 − 4y + 4 = y 2 + 4y + 4; x2 = 8y The diagram makes it clear that the vertex is (0, 0) and the focal length is a = 2. Hence the length of the latus rectum is 4a = 8. Copyright © 2000 - 2010 Yimin Math Centre (www.yiminmathcentre.com)

Year 11 Term 3 Week 1 Homework 11.2.4

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The four standard positions of the parabola

The equation of every parabola whose vertex as the origin and whose axis is vertical or horizontal can be put into exactly one of the four form as shown: x2 = 4ay

x2 = −4ay

y 2 = 4ax

y 2 = −4ax

Exercise 11.2.2 Applying the method outlined in the previous question, use definition of a parabola to derive the equations of: 1. the parabola with focus (0, 3) and directrix y + 3 = 0

2. the parabola with focus (0, −2) and directrix y − 2 = 0

3. the parabola with focus (2, 0) and directrix x + 2 = 0

4. the parabola with focus (− 43 , 0) and directrix x −

4 3

=0

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