Preliminary Mathematics (2U)

Transcription

1 Preliminary Mathematics (2U) Week 2 Student Name:. Class code:.. Teacher name:.. DUX

2 Week 2 Theory A Parabola: 2 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.

3 3 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. 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 x 2 + (y a) 2 = (y + a) 2 x 2 + y 2 2ay + a 2 = y 2 + 2ay + a 2 x 2 = 4ay

4 Case Two (Example): 4 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. 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 x 2 + (y + a) 2 = (y a) 2 x 2 + y 2 + 2ay + a 2 = y 2 2ay + a 2 x 2 = 4ay

5 Case Three (Example): 5 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. Given the distance of FP = AP, so use the distance formula: FP = AP (x a) 2 + (y 0) 2 = (x a) 2 + (y y) 2 (x a) 2 + y 2 = (x + a) x 2 2ax + a 2 + y 2 = x 2 + 2ax + a 2 y 2 = 4ax

6 Case Four (Example): 6 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. Given the distance of FP = AP, so use the distance formula: FP = AP (x a) 2 + (y 0) 2 = (x a) 2 + (y y) 2 (x + a) 2 + y 2 = (x a) x 2 + 2ax + a 2 + y 2 = x 2 2ax + a 2 y 2 = 4ax

7 Example: 7 Sketch the parabola with equation x 2 = 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. x 2 = 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.

8 8 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. Given the distance of FP = AP, so use the distance formula: 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 + y 2 2(k + a)y + (k + a) 2 = y 2 2(k a)y + (k a) 2 (x h) 2 + y 2 2ky 2ay + k 2 + 2ak + a 2 = y 2 2ky + 2ay + k 2 2ak + a 2 (x h) 2 = 4ay 4ak (x h) 2 = 4a(y k)

9 Case Two (Example): 9 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. Given the distance of FP = AP, so use the distance formula: 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 + y 2 2(k a)y + (k a) 2 = y 2 2(k + a)y + (k + a) 2 (x h) 2 + y 2 2ky + 2ay + k 2 2ak + a 2 = y 2 2ky 2ay + k 2 + 2ak + a 2 (x h) 2 = 4ay + 4ak (x h) 2 = 4a(y k)

10 Case Three (Example): 10 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. Given the distance of FP = AP, so use the distance formula: 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 x 2 2(h + a)x + (h + a) 2 + (y k) 2 = x 2 2(h a)x + (h a) 2 x 2 2hx 2ax + h 2 + 2ah + a 2 + (y k) 2 = x 2 2hx + 2ax + h 2 2ah + a 2 (y k) 2 = 4ax 4ah (y k) 2 = 4a(x h)

11 Case Four (Example): 11 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. Given the distance of FP = AP, so use the distance formula: 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 x 2 2(h a)x + (h a) 2 + (y k) 2 = x 2 2(h + a)x + (h + a) 2 x 2 2hx + 2ax + h 2 2ah + a 2 + (y k) 2 = x 2 2hx 2ax + h 2 + 2ah + a 2 (y k) 2 = 4ax + 4ah (y k) 2 = 4a(x h)

12 Example: 12 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. Firstly, sketch a diagram. Given the distance of FP = AP, so use the distance formula: 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 + y 2 4y + 4 = y y + 36 (x 3) 2 = 16y + 32 (x 3) 2 = 16(y + 2)

13 Example: 13 For the parabola with equation 2y 2 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. (i) 2y 2 8y + x + 9 = 0 y 2 4y x = 0 y 2 4y + ( 2) 2 = 1 2 x ( 2)2 (y 2) 2 = 1 2 x 1 2 (y 2) 2 = 1 (x + 1) 2 The focal length: 4a = 1 2 a = 1 8 The vertex is ( 1, 2) and the focus is ( 1 1 8, 2). (ii) The equation of the directrix is x = 7 and the axis of symmetry is y = 2. 8 (iii)

14 Example: 14 Find the equation of the parabola with vertex (2, 3) and the equation of the directrix is x = 3. 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)

15 DUX Week 2 Homework 15 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) x 2 = 4y b) x 2 = y c) x 2 = 0.5y d) y 2 = 4x e) y 2 = 1 6 x f) (x + 1) 2 = 4(y 1) g) (x 3) 2 = 1 16 y h) (y 2) 2 = 8(x + 1) i) (y + 3) 2 = 1 (x 2) 2

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17 17 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.

18 18 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) x 2 2x y + 4 = 0 b) y 2 6y + 8x + 1 = 0 c) 2x 2 8x + y + 6 = 0 d) 4y y x + 38 = 0

19 19 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 unit and the axis of symmetry is the x-axis. 2 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