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Analytical Geometry XII - Standard Mathematics. For the ellipse If a  b, then AA’ = 2a is major axis BB’ = 2b is minor axis Focus S(  ae, 0) Directrix.

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Presentation on theme: "Analytical Geometry XII - Standard Mathematics. For the ellipse If a  b, then AA’ = 2a is major axis BB’ = 2b is minor axis Focus S(  ae, 0) Directrix."— Presentation transcript:

1 Analytical Geometry XII - Standard Mathematics

2 For the ellipse If a  b, then AA’ = 2a is major axis BB’ = 2b is minor axis Focus S(  ae, 0) Directrix DD’ is x =  a/e Center C(0,0) Eccentricity is given by b 2 = a 2 (1 – e 2 ) SS’AA’ B B’ C D D’

3 For the ellipse S S’ AA’ B B’ C DD’ D1D1 D1’D1’ If a > b, then AA’ = 2b is minor axis BB’ = 2a is major axis Focus S(0,  ae ) Directrix DD’ is y =  a/e Center C(0,0) Eccentricity is given by b 2 = a 2 (1 – e 2 )

4 For the hyperbola AA’ = 2a is transverse axis A(a,0) and A’(-a,0) are vertices BB’ = 2b is conjugate axis Focus S(  ae, 0) Directrix DD’ is x =  a/e Center C(0,0) Eccentricity is given by b 2 = a 2 (e 2 –1) AA’SS’C D D’ D1D1 D1’D1’

5 Find the vertex, focus the latus rectum, axis and the directrix of the parabola y 2 = 8x The equation of the parabola is y 2 = 8x  4a = 8 a = 2 Vertex = (0,0) Focus = S(2,0) Latus rectum LL’ = 4a = 4  2 = 8 Axis of the parabola is y = 0 Directrix is x = -a ie) x = -2 S(2,0)x y x=-2

6 Find the axis, focus, latus rectum,equation of LR, vertex and the directrix of the parabola y 2 – 8x – 2y + 17 = 0 The equation of the parabola is y 2 – 8x – 2y + 17 = 0 y 2 – 2y = 8x – 17 y 2 – 2y + 1 = 8x – (y – 1) 2 = 8x – 16 (y – 1) 2 = 8(x – 2) Y 2 = 8X where X = x – 2 and Y = y – 1  4a = 8 a = 2

7 Referred to X, Y axis Axis – X-axis Y = 0 Referred to X, Y axis y – 1 = 0 y = 1 Focus = (a, 0) (2, 0) Focus = (a, 0) = (2, 0) x – 2 = 2, y – 1 = 0 x = = 4, y = 1 Focus = (4, 1) Length of latus rectum = 4a = 4  2 = 8 Length of latus rectum = 4a = 4  2 = 8 equation of latus rectum is X = a, ie) X = 2 Equation of latus rectum is x – 2 = 2, ie) x = 4

8 Directrix is X = – a ie) X = – 2 Referred to X, Y axesReferred to x, y axes Vertex = (0, 0) X = 0, Y = 0 Vertex = (0, 0) x – 2 = 0, y – 1 = 0 x = 2, y = 1 Vertex = (2, 1) Directrix is x – 2 = –2 x = – ie) x = 0 S(4,1)X Y x=0 x y C(2,1)

9 Find the centre, vertices, foci, eccentricity, latus rectum, and directrices of the ellipse 3x 2 + 4y 2 = 12 The equation of the ellipse is 3x 2 + 4y 2 = 12  a 2 = 4, b 2 = 3 and a  b a = 2, b =  3 b 2 = a 2 (1 – e 2 ) 3 = 4 (1 – e 2 ) (1 – e 2 ) = ¾ e 2 = 1 – ¾ = ¼ e = ½

10 Centre (0,0) Vertices (  a, 0) = (  2, 0) Foci = (  ae, 0) = (  2  ½, 0) = (  1, 0) Eccentricity e = ½ Latus rectum LL’ = Directrices x = Directrices x =  4

11 Find the centre, vertices, foci, eccentricity, latus rectum, and directrices of the ellipse 25x 2 + 9y 2 = 225. The equation of the ellipse is 25x 2 + 9y 2 = 225  a 2 = 9, b 2 = 25 and b  a a = 3, b = 5 a 2 = b 2 (1 – e 2 ) 9 = 25 (1 – e 2 ) (1 – e 2 ) = 9/25 e 2 = 1 – 9/25 = 16/25 e = 4/5

12 Centre (0,0) Vertices (0,  b) = (0,  5) Foci = (0,  be) = (0,  5  4/5) = (0,  4) Eccentricity e = 4/5 Latus rectum LL’ = Directrices y =

13 Equation of the ellipse is 36x 2 + 4y 2 – 72x + 32y – 44 = 0 36x 2 – 72x + 4y y = 44 36(x 2 – 2x) + 4(y 2 + 8y) = 44 36(x 2 – 2x + 1 – 1) + 4(y 2 + 8y + 16 – 16) = 44 36(x – 1) 2 – (y + 4) 2 – 64 = 44 36(x – 1) 2 + 4(y + 4) 2 = (x – 1) 2 + 4(y + 4) 2 = 144 Find the eccentricity, centre, foci vertices and directrices of the ellipse 36x 2 + 4y 2 – 72x + 32y – 44 = 0

14 Where X = x – 1, Y = y + 4  a 2 = 36, b 2 = 4 a = 6, b = 2 b 2 = a 2 (1 – e 2 ) 4 = 36 (1 – e 2 ) (1 – e 2 ) = 4/36 e 2 = 1 – 4/36 = 32/36 Eccentricity e =  32/6 e = 4  2/

15 Equation of the ellipse is 16x 2 + 9y x – 36y = 92 16x x + 9y 2 – 36y = 92 16(x 2 + 2x) + 9(y 2 – 4y) = 92 16(x 2 + 2x + 1 – 1) + 9(y 2 – 4y + 4 – 4) = 92 16(x + 1) 2 – (y – 2) 2 – 36 = 92 16(x + 1) 2 + 9(y – 2) 2 = (x + 1) 2 + 9(y – 2) 2 = 144 Find the eccentricity, centre, foci vertices and directrices of the ellipse 16x 2 + 9y x – 36y = 92

16 Where X = x + 1, Y = y – 2  a 2 = 16, b 2 = 9 a = 4, b = 3 b 2 = a 2 (1 – e 2 ) 9 = 16 (1 – e 2 ) (1 – e 2 ) = 9/16 e 2 = 1 – 9/16 = 7/16 Eccentricity e =  7/4

17 Referred to (X,Y) Referred to (x,y) Centre (0,0) X = 0, Y = 0 X = x + 1,Y = y – 2  x = X – 1,y = Y + 2 x = 0 – 1, y = x = –1, y = 2 Centre = C(–1, 2) Vertex (  a, 0) = (  4,0) X =  4, Y = 0 x = X – 1, y = Y + 2 x =  4 + 2, y = x = 6, –2, y = 2 Vertices are A(6,2) and A’(–2,2)

18 Foci (  ae, 0) = (  4  7/4,0) = (  7, 0) X =  7, Y = 0 x = X – 1, y = Y + 2 x =  7 – 1, y = x =  7–1, –  7–1, y = 2 Foci are (  7–1,2) and (–  7–1,2) Latus rectum LL’= Latus rectum = 9/2 Directrix of the ellipse is x = X – 1 =  16/  7 – 1 = 16/  7–1, –16/  7–1 Directrices of the parabola are x = 16/  7–1 and x = –16/  7–1

19 Equation of the hyperbola is x 2 – 4y 2 + 6x + 16y – 11 = 0 (x 2 + 6x) – 4(y 2 – 4y) = 11 (x 2 + 6x + 9 – 9) – 4(y 2 – 4y + 4 – 4) = 11 (x + 3) 2 – 9 – 4(y – 2) = 11 (x + 3) 2 – 4(y – 2) 2 = – 16 (x + 3) 2 – 4(y – 2) 2 = 4 Find the eccentricity, centre, foci vertices and directrices of the hyperbola x 2 – 4y 2 + 6x + 16y – 11 = 0

20 Where X = x + 3, Y = y – 2  a 2 = 4, b 2 = 1 and a  b a = 2, b = 1 b 2 = a 2 (e 2 – 1) 1 = 4 (e 2 – 1) (e 2 – 1) = 1/4 e 2 = 1 + 1/4 = 5/4 Eccentricity e =  5/2

21 Referred to (X,Y) Referred to (x,y) Centre (0,0) X = 0, Y = 0 X = x + 3,Y = y – 2  x = X – 3,y = Y + 2 x = 0 – 3, y = x = –3, y = 2 Centre = C(–3, 2) Vertex (  a, 0) = (  2,0) X =  2, Y = 0 x = X – 1, y = Y + 2 x =  2 + 2, y = x = 4, 0, y = 2 Vertices are A(4,2) and A’(0,2)

22 Foci (  ae, 0) = (  2  5/2,0) = (  5, 0) X =  5, Y = 0 x = X – 3, y = Y + 2 x =  5 – 3, y = x =  5–3, –  5–3, y = 2 Foci are (  5–3,2) and (–  5–3,2) Latus rectum LL’= Latus rectum = 1 Directrix of the ellipse is x = X – 3 =  4/  5 – 3 = 4/  5–3, –4/  5–3 Directrices of the parabola are x = 4/  5–3 and x = –4/  5–3

23 Equation of the ellipse is Find the centre, vertices, foci, eccentricity, latus rectum, and directrices of the ellipse Where X = x + 1, Y = y – 2  a 2 = 9, b 2 = 5 and a  b a = 3, b =  5 b 2 = a 2 (1 – e 2 ) 5 = 9 (1 – e 2 ) (1 – e 2 ) = 5/9 e 2 = 1 – 5/9 = 4/9 e = 2/3

24 Referred to (X,Y) Referred to (x,y) Centre (0,0) X = 0, Y = 0 X = x + 1,Y = y – 2  x = X – 1,y = Y + 2 x = 0 – 1, y = x = - 1, y = 2 Vertex = (-1, 2) Vertex (  a, 0) = (  3,0) X =  3, Y = 0 x = X – 1, y = Y + 2 x =  3 + 2, y = x = 5, -1 y = 2 Vertices are (5,2) and (-1,2)

25 Foci (  ae, 0) = (  3  2/3,0) = (  2, 0) X =  2, Y = 0 x = X – 1, y = Y + 2 x =  2 – 1, y = x = 1, -3 y = 2 Foci are (1,2) and (-3,2) Latus rectum LL’= Latus rectum = 10/3 Directrix of the ellipse is x = X – 1 =  9/2 – 1 = 9/2 – 1, -9/2 – 1 = 7/2, -11/2 Directrix of the parabola is x = 7/2 and x = -11/2

26 An arch is in the form of a semi ellipse whose span is 48feet wide. The height of the arch is 20feet. How wide is the arch at a height of 10feet above the base? Take the mid point of the base as the centre C(0, 0) Width of the base = 48ft = AA’ = 2a  CA = 24ft The vertices are A(24, 0), A’(-24, 0) Height of the arch BC = 20 = b  a = 24, b = 20 The equation of the ellipse is C(0,0) A(24,0) P(x,10) A’(-24,0) Q B R

27 Let x 1 be the distance between the pole whose height is 10feet and the centre The point (x 1, 10) is a point on the ellipse Width of the arch at a height of 10feet = 2x 1

28 Find the equation of the rectangular hyperbola which has one of its asymptotes the line x + 2y – 5 = 0 and passes through the points (6, 0) and (–3, 0) Since the asymptotes are perpendicular, their equations arex + 2y – 5 = 0 and 2x – y + k = 0 The equation of the rectangular hyperbola differ only by its constant  The equation of the rectangular hyperbola is (x + 2y – 5)(2x – y + k) + m = 0 It passes through the points (6, 0) and (–3, 0) (6 + 0 – 5)(12 – 0 + k) + m = 0 (1)(12 + k) + m = 0 k + m = –12 ………..(1)and

29 It passes through the point (–3, 0) (–3 + 0 – 5)(–6 – 0 + k) + m = 0 (–8)(–6 + k) + m = 0 48 – 8k + m = 0 8k – m = 48………..(2) (1)  k + m = –12 (2)  8k – m = 48 (1)+(2)  9k = 36 k = 4 Sub k = 4 in (1) 4 + m = –12 m = – 12 – 4 = –16 The equation of the rectangular hyperbola is (x + 2y – 5)(2x – y + 4) – 16 = 0

30 Find the equation of the hyperbola if its asymptotes parallel to x + 2y – 12 = 0 and x – 2y + 8 = 0, (2, 4) is the centre of the hyperbola and it passes through(2, 0) Since the asymptotes are parallel to the lines x + 2y – 12 = 0 and x – 2y + 8 = 0, The equation of the asymptotes are in the form x + 2y + k = 0 and x – 2y + m = 0 Since the asymptotes pass through the centre (2,4) k = k = 0 k = – 10 and 2 – 8 + m = 0 – 6 + m = 0 m = 6 The equations of the asymptotes are x + 2y – 10 = 0 x – 2y + 6 = 0

31  The equation of the hyperbola is (x + 2y – 10)(x – 2y + 6) + l = 0 It passes through the point (2, 0) (2 + 0 – 10)(2 – 0 + 6) + l = 0 (–8)(8) + l = 0 – 64 + l = 0 l = 64  The equation of the hyperbola is (x + 2y – 10)(x – 2y + 6) + 64 = 0

32 Find the equation of the rectangular hyperbola which has its centre at (2, 1), one of its asymptotes 3x – y – 5 = 0 and which passes through the point (1, –1). Equation of the asymptote is 3x – y – 5 = 0 Equation of the other asymptote is x + 3y + k = 0 It passes through the centre (2, 1) k = 0 k + 5 = 0 k = –5 Equation of the other asymptote is x + 3y – 5 = 0 Equation of the rectangular hyperbola is (3x – y – 5)(x + 3y – 5) + m = 0

33 It passes through the point (1, – 1) (3 + 1 – 5)(1 – 3 – 5) + m = 0 (–1)(–7) + m = m = 0 m = –7  Equation of the rectangular hyperbola is (3x – y – 5)(x + 3y – 5) – 7 = 0

34 Find the equation of the rectangular hyperbola which has for one of its asymptotes the line x + 2y – 5 = 0 and passes through the points (6, 0) and (–3, 0). Equation of the asymptote is x + 2y – 5 = 0 Equation of the other asymptote is 2x – y + k = 0 Combined Equation of the asymptotes is (x + 2y – 5) (2x – y + k) = 0 Equation of the rectangular hyperbola is (x + 2y – 5) (2x – y + k) + m = 0 It passes through the points (6, 0) and (–3, 0) (6 + 0 – 5)(12 – 0 + k) + m = 0 (1)(12 + k) + m = 0 k + m = – 12 ……….(1)

35 It passes through the point (–3, 0) (–3 + 0 – 5)(– 6 – 0 + k) + m = 0 (1 – 8)(–6 + k) + m = 0 48 – 8k + m = 0 –8k + m = –48……….(2) (1)  k + m = – 12 (2)  –8k + m = – 48 (1) – (2)  9k = 36 k = 4 Sub k = 4 in eqn (1) 4 + m = –12 m = – 16 The equation of the rectangular hyperbola is (x + 2y – 5) (2x – y + 4) – 16 = 0


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