Download presentation

Presentation is loading. Please wait.

1
**XII - Standard Mathematics**

Analytical Geometry PREPARED BY: R.RAJENDRAN. M.A., M. Sc., M. Ed., K.C.SANKARALINGA NADAR HR. SEC. SCHOOL, CHENNAI-21

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 b2 = a2(1 – e2) D •B A’• •S’ •C •S •A •B’ D’

3
**For the ellipse 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 b2 = a2(1 – e2) D D’ •B •S A’• •C •A •S’ •B’ D1 D1’

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 b2 = a2(e2 –1) D1 D •S’ •A’ •C •A •S D’ D1’

5
**The equation of the parabola is y2 = 8x 4a = 8 a = 2**

Find the vertex, focus the latus rectum, axis and the directrix of the parabola y2 = 8x The equation of the parabola is y2 = 8x 4a = a = 2 y x=-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

6
Find the axis, focus, latus rectum,equation of LR, vertex and the directrix of the parabola y2 – 8x – 2y + 17 = 0 The equation of the parabola is y2 – 8x – 2y + 17 = 0 y2 – 2y = 8x – 17 y2 – 2y + 1 = 8x – (y – 1)2 = 8x – 16 (y – 1)2 = 8(x – 2) Y2 = 8X where X = x – 2 and Y = y – 1 4a = a = 2

7
Referred to X, Y axis Referred to X, Y axis Axis – X-axis Y = 0 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
**Referred to X, Y axes Referred to x, y axes Vertex = (0, 0)**

Directrix is X = – a ie) X = – 2 Directrix is x – 2 = –2 x = – 2 + 2 ie) x = 0 y Y x=0 C(2,1)• •S(4,1) X x

9
Find the centre, vertices, foci, eccentricity, latus rectum, and directrices of the ellipse 3x2 + 4y2 = 12 The equation of the ellipse is 3x2 + 4y2 = 12 b2 = a2 (1 – e2) 3 = 4 (1 – e2) (1 – e2) = ¾ e2 = 1 – ¾ = ¼ e = ½ a2 = 4, b2 = 3 and a b a = 2 , b = 3

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 25x2 + 9y2 = 225. The equation of the ellipse is 25x2 + 9y2 = 225 a2 = b2 (1 – e2) 9 = 25 (1 – e2) (1 – e2) = 9/25 e2 = 1 – 9/25 = 16/25 e = 4/5 a2 = 9, b2 = 25 and b a a = 3 , b = 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 = Directrices y =

13
Find the eccentricity, centre, foci vertices and directrices of the ellipse 36x2 + 4y2 – 72x + 32y – 44 = 0 Equation of the ellipse is 36x2 + 4y2 – 72x + 32y – 44 = 0 36x2 – 72x + 4y2 + 32y = 44 36(x2 – 2x) + 4(y2 + 8y) = 44 36(x2 – 2x + 1 – 1) + 4(y2 + 8y + 16 – 16) = 44 36(x – 1)2 – (y + 4)2 – 64 = 44 36(x – 1)2 + 4(y + 4)2 = 36(x – 1)2 + 4(y + 4)2 = 144

14
b2 = a2 (1 – e2) 4 = 36 (1 – e2) (1 – e2) = 4/36 e2 = 1 – 4/36 = 32/36 Eccentricity e = 32/6 e = 4 2/ Where X = x – 1, Y = y + 4 a2 = 36, b2 = 4 a = 6, b = 2

15
Find the eccentricity, centre, foci vertices and directrices of the ellipse 16x2 + 9y2 + 32x – 36y = 92 Equation of the ellipse is 16x2 + 9y2 + 32x – 36y = 92 16x2 + 32x + 9y2 – 36y = 92 16(x2 + 2x) + 9(y2 – 4y) = 92 16(x2 + 2x + 1 – 1) + 9(y2 – 4y + 4 – 4) = 92 16(x + 1)2 – (y – 2)2 – 36 = 92 16(x + 1)2 + 9(y – 2)2 = 16(x + 1)2 + 9(y – 2)2 = 144

16
b2 = a2 (1 – e2) 9 = 16 (1 – e2) (1 – e2) = 9/16 e2 = 1 – 9/16 = 7/16 Eccentricity e = 7/4 Where X = x + 1, Y = y – 2 a2 = 16, b2 = 9 a = 4 , b = 3

17
**Referred to (X,Y) Referred to (x,y)**

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

18
x = X – 1 , y = Y + 2 x = 7 – 1 , y = 0 + 2 x = 7–1, – 7–1, y = 2 Foci are (7–1,2) and (– 7–1,2) Foci (ae, 0) = (47/4,0) = (7, 0) X = 7, Y = 0 Latus rectum LL’= Latus rectum = 9/2 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 Directrix of the ellipse is

19
Find the eccentricity, centre, foci vertices and directrices of the hyperbola x2 – 4y2 + 6x + 16y – 11 = 0 Equation of the hyperbola is x2 – 4y2 + 6x + 16y – 11 = 0 (x2 + 6x) – 4(y2 – 4y) = 11 (x2 + 6x + 9 – 9) – 4(y2 – 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

20
b2 = a2 (e2 – 1) 1 = 4 (e2 – 1) (e2 – 1) = 1/4 e2 = 1 + 1/4 = 5/4 Eccentricity e = 5/2 Where X = x + 3, Y = y – 2 a2 = 4, b2 = 1 and a b a = 2 , b = 1

21
**Referred to (X,Y) Referred to (x,y)**

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

22
x = X – 3 , y = Y + 2 x = 5 – 3 , y = 0 + 2 x = 5–3, – 5–3, y = 2 Foci are (5–3,2) and (–5–3,2) Foci (ae, 0) = (25/2,0) = (5, 0) X = 5, Y = 0 Latus rectum LL’= Latus rectum = 1 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 Directrix of the ellipse is

23
**Find the centre, vertices, foci, eccentricity, latus rectum, and directrices of the ellipse**

Equation of the ellipse is b2 = a2 (1 – e2) 5 = 9 (1 – e2) (1 – e2) = 5/9 e2 = 1 – 5/9 = 4/9 e = 2/3 Where X = x + 1, Y = y – 2 a2 = 9, b2 = 5 and a b a = 3 , b = 5

24
**Referred to (X,Y) Referred to (x,y)**

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

25
x = X – 1 , y = Y + 2 x = 2 – 1 , y = 0 + 2 x = 1, y = 2 Foci are (1,2) and (-3,2) Foci (ae, 0) = (32/3,0) = (2, 0) X = 2, Y = 0 Latus rectum LL’= Latus rectum = 10/3 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 Directrix of the ellipse is

26
**Take the mid point of the base as the centre C(0, 0) **

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 B Q P(x,10) A(24,0) A’(-24,0) C(0,0) R

27
**Let x1 be the distance between the pole whose height is 10feet and the centre**

The point (x1, 10) is a point on the ellipse Width of the arch at a height of 10feet = 2x1

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 are x + 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 = 0 10 + 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 (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
**Equation of the rectangular hyperbola is (3x – y – 5)(x + 3y – 5) + m = 0**

It passes through the point (1, – 1) (3 + 1 – 5)(1 – 3 – 5) + m = 0 (–1)(–7) + m = 0 7 + 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)**

(– – 5)(– 6 – 0 + k) + m = 0 (1 – 8)(–6 + k) + m = 0 48 – 8k + m = 0 –8k + m = –48……….(2) k + m = – 12 –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

Similar presentations

OK

OHHS Pre-Calculus Mr. J. Focht. 8.3 Hyperbolas Geometry of a Hyperbola Translations of Hyperbolas Eccentricity 8.3.

OHHS Pre-Calculus Mr. J. Focht. 8.3 Hyperbolas Geometry of a Hyperbola Translations of Hyperbolas Eccentricity 8.3.

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on share trading in india Ppt on nutrition in human beings Ppt on solar energy usage Ppt on natural resources for class 10 Ppt on water quality standards Ppt on airport site selection Ppt on reflection of sound Ppt on conventional energy sources Ppt on fire extinguisher types and uses Download ppt on statistics for class 10th