XII - Standard Mathematics Analytical Geometry 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) D •B A’• •S’ •C •S •A •B’ D’

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’

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’

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 = 8 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

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 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 = 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

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

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

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

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

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 =

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 – 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

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

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 – 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

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

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

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) = (47/4,0) = (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

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)2 + 16 = 11 (x + 3)2 – 4(y – 2)2 = 11 + 9 – 16 (x + 3)2 – 4(y – 2)2 = 4

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

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

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) = (25/2,0) = (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

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

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, -1 y = 2 Vertices are (5,2) and (-1,2) Vertex (a, 0) = (3,0) X = 3, 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) Foci (ae, 0) = (32/3,0) = (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

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

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

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 = 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

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) 2 + 8 + 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

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

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)  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