analytical geometry

Analytical Geometry

XII - Standard

Mathematics

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

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)

1b

y

a

x2

2

2

2

•S•S’ •AA’•

•B

•B’

•C

D

D’

For the ellipse 1a

y

b

x2

2

2

2

•S

•S’

•AA’•

•B

•B’

•C

D D’

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)


b2 = a2(1 – e2)

For the hyperbola 1b

y

a

x2

2

2

2

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)


b2 = a2(e2 –1)

•A•A’ •S•S’ •C

D

D’

D1

D1’

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 = 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

yx=-2

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 – 17 + 1

(y – 1)2 = 8x – 16

(y – 1)2 = 8(x – 2)

Y2 = 8X where X = x – 2 and Y = y – 1

4a = 8 a = 2

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 = 2 + 2 = 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

Directrix is X = – a

ie) X = – 2

Referred to X, Y axes Referred 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 = – 2 + 2

ie) x = 0

•S(4,1) X

Yx=0

x

y

C(2,1)•

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

112

4y

12

3x 22

13

y

4

x 22

a2 = 4, b2 = 3 and a b a = 2 , b = 3

b2 = a2 (1 – e2)

3 = 4 (1 – e2)

(1 – e2) = ¾

e2 = 1 – ¾ = ¼

e = ½

Centre (0,0)

Vertices ( a, 0) = (2 , 0)

Foci = ( ae, 0) = ( 2 ½, 0) = ( 1, 0)

Eccentricity e = ½

Latus rectum LL’ = 32

32

a

b2 2

Directrices x = 42/1

2

e

a

Directrices x = 4

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

1225

9y

225

25x 22

125

y

9

x 22

a2 = 9, b2 = 25 and b a a = 3 , b = 5

a2 = b2 (1 – e2)9 = 25 (1 – e2)(1 – e2) = 9/25 e2 = 1 – 9/25 = 16/25 e = 4/5

Centre (0,0)

Vertices (0, b) = (0, 5)

Foci = (0, be) = (0, 5 4/5) = (0, 4)

Eccentricity e = 4/5

Latus rectum LL’ = 5

18

5

92

b

a2 2

Directrices y = 4

25

5/4

5

e

b

Directrices y = 4

25

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 – 36 + 4(y + 4)2 – 64 = 44

36(x – 1)2 + 4(y + 4)2 = 44 + 36 + 64

36(x – 1)2 + 4(y + 4)2 = 144

Find the eccentricity, centre, foci vertices and directrices of the ellipse 36x2 + 4y2 – 72x + 32y – 44 = 0

1144

4)4(y

144

1)36(x 22

136

Y

4

X 22

Where X = x – 1, Y = y + 4

a2 = 36, b2 = 4

a = 6, b = 2

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/


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 – 16 + 9(y – 2)2 – 36 = 92

16(x + 1)2 + 9(y – 2)2 = 92 + 16 + 36

16(x + 1)2 + 9(y – 2)2 = 144

Find the eccentricity, centre, foci vertices and directrices of the ellipse 16x2 + 9y2 + 32x – 36y = 92

1144

2)9(y

144

1)16(x 22

116

Y

9

X 22

Where X = x + 1, Y = y – 2

a2 = 16, b2 = 9

a = 4 , b = 3

b2 = a2 (1 – e2)

9 = 16 (1 – e2)

(1 – e2) = 9/16

e2 = 1 – 9/16

= 7/16

Eccentricity

e = 7/4

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 = 0 + 2

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 = 0 + 2

x = 6, –2, y = 2

Vertices are A(6,2) and A’(–2,2)

Foci (ae, 0) = (47/4,0)

= (7, 0)

X = 7, Y = 0

x = X – 1 , y = Y + 2 x = 7 – 1 , y = 0 + 2 x = 7–1, – 7–1, y = 2Foci 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

2

9

4

92

a

b2 2

7

164/7

4

e

aX

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)2 + 16 = 11

(x + 3)2 – 4(y – 2)2 = 11 + 9 – 16

(x + 3)2 – 4(y – 2)2 = 4

Find the eccentricity, centre, foci vertices and directrices of the hyperbola x2 – 4y2 + 6x + 16y – 11 = 0

14

2)4(y

4

3)(x 22

11

2)(y

4

3)(x 22

11

Y

4

X 22

Where X = x + 3, Y = y – 2

a2 = 4, b2 = 1

and a b a = 2 , b = 1

b2 = a2 (e2 – 1)1 = 4 (e2 – 1)(e2 – 1) = 1/4 e2 = 1 + 1/4 = 5/4 Eccentricity

e = 5/2


Centre (0,0)

X = 0 , Y = 0

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)

Vertex (a, 0) = (2,0)

X = 2, 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)

Foci (ae, 0) = (25/2,0)

= (5, 0)

X = 5, Y = 0

x = X – 3 , y = Y + 2 x = 5 – 3 , y = 0 + 2 x = 5–3, – 5–3, y = 2Foci are (5–3,2) and (–5–3,2)

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

12

12

a

b2 2

5

42/5

2

e

aX


Find the centre, vertices, foci, eccentricity, latus

rectum, and directrices of the ellipse 15

2)-(y

9

1)(x 22

15

2)-(y

9

1)(x 22

15

Y

9

X 22

Where X = x + 1, Y = y – 2

a2 = 9, b2 = 5 and a b a = 3 , b = 5

b2 = a2 (1 – e2)5 = 9 (1 – e2)(1 – e2) = 5/9 e2 = 1 – 5/9 = 4/9 e = 2/3


Centre (0,0)

X = 0 , Y = 0

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)

Vertex (a, 0) = (3,0)

X = 3, Y = 0

x = X – 1 , y = Y + 2

x = 3 + 2, y = 0 + 2

x = 5, -1 y = 2

Vertices are (5,2) and (-1,2)

Foci (ae, 0) = (32/3,0)

= (2, 0)

X = 2, Y = 0

x = X – 1 , y = Y + 2

x = 2 – 1 , y = 0 + 2

x = 1, -3 y = 2

Foci are (1,2) and (-3,2)

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

3

10

3

52

a

b2 2

2

9

3/2

3

e

ax

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

120

y

24

x2

2

2

2

R

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

120

10

24

x2

2

2

21

1400

100

576

x 21

14

1

576

x 21

4

3

4

11

576

x 21

5764

3x 2

1

242

3x1

feet 312x1

Width of the arch at a height of 10feet = 2x1

feet 324

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

It passes through the point (–3, 0)(–3 + 0 – 5)(–6 – 0 + k) + m = 0(–8)(–6 + k) + m = 0 48 – 8k + m = 08k – m = 48………..(2)(1) k + m = –12(2) 8k – m = 48(1)+(2) 9k = 36

k = 4Sub 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

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 formx + 2y + k = 0 and x – 2y + m = 0

Since the asymptotes pass through the centre (2,4)2 + 8 + k = 010 + k = 0 k = – 10 and2 – 8 + m = 0– 6 + m = 0 m = 6

The equations of the asymptotes are

x + 2y – 10 = 0

x – 2y + 6 = 0

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

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)

2 + 3 + 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

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

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)

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

analytical geometry

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