FIND VERTEX FOCUS EQUATION OF DIRECTRIX OF HYPERBOLA

Details of hyperbola whose center is (0, 0)

Symmetric about x- axis

Center :

C (0, 0)

Foci :

F1 (ae, 0) F2 (-ae, 0)

Vertices on transverse axis  :

A (a, 0) A' (-a, 0)

Vertices on conjugate axis  :

B (0, b) B' (0, -b)

Equation of directrices :

x  =  ± (a/e)

Symmetric about y- axis

Center :

C (0, 0)

Foci :

F1 (0, ae) F2 (0, -ae)

Vertices on transverse axis  :

A (0, a) A' (0, -a)

Vertices on conjugate axis  :

B (b, 0) B' (-b, 0)

Equation of directrices :

y  =  ± (a/e)

Details of hyperbola whose center is (0, 0)

Symmetric about x- axis

Center :

C (h, k)

Foci :

F1 (h + c, k ) F2 (h - c, k )

Vertices on major axis  :

A (h + a, k) A' (h - a, k)

Equation of directrices :

x  =  h ± (a/e)

Symmetric about y- axis

Center :

C (h, k)

Foci :

F(h, k + c) F(h, k - c)

Vertices on major axis  :

A (h, k + a ) A' (h, k - a)

Equation of directrices :

y  = k ± (a/e)

Note :

e  =  √[1 + (b2/a2)]

b=  a2(e2 - 1)

Question :

Identify the type of conic and find centre, foci, vertices, and directrices of each of the following:

(i)  [(x + 3)2/225] - [(y - 4)2/64] = 1

Solution :

The given conic represents the " Hyperbola "

The given ellipse is symmetric about x - axis.

a2  =  225 and b =  64

a  =  15 and b  =  8

c2  =  a2 + b2

c2  =  225 + 64

c2  =  289

c  =  17

e = c/a  =  17/15

Center :

C (h, k)  ==>  (-3, 4)

Foci :

F1 (h + c, k ) F2 (h - c, k )

F1 (-3 + 17, 4) F2 (-3 - 17, 4 )

F1 (14, 4) F2 (-20, 4 )

Vertices on transverse axis  :

A (h + a, k) A' (h - a, k)

A (-3 + 15, 4) A' (-3 - 15, 4)

A (12, 4) A' (-18, 4)

Equation of directrices :

x  =  h ± (a/e)

x  =  -3 ± (15/(17/15))

x  =  -3 ± (225/17)

  x  =  -3 + (225/17)

  x  =  (-51 + 225)/17

  x  =  174/17

  x  =  -3 - (225/17)

  x  =  (-51 - 225)/17

  x  =  -276/17

(ii)  [(y - 2)2/25] - [(x + 1)2/16] = 1

Solution :

The given conic represents the " Hyperbola "

The given ellipse is symmetric about y - axis.

a2  =  25 and b =  16

a  =  5 and b  =  4

c2  =  a2 + b2

c2  =  25 + 16

c2  =  41

c  =  √41

e = c/a  =  √41/5

Center :

C (h, k)  ==>  (-1, 2)

Foci :

F1 (h, k + c) F2 (h, k - c)

F1 (-1, 2 + √41) F2 (-1, 2 - √41)

Vertices on transverse axis  :

A (h, k + a) A' (h, k - a)

A (-1, 2 + 5) A' (-1, 2 - 5)

A (-1, 7) A' (-1, -3)

Equation of directrices :

y  =  k ± (a/e)

y  =  2 ± (5/(√41/5))

y  =  2 ± (25/√41)

y  =  2 + (25/√41)

  y  =  2 - (25/√41)

(v) 18x2 + 12y2 − 144x + 48y + 120 = 0

Solution :

18x2 + 12y2 − 144x + 48y + 120 = 0

18x2 − 144x + 12y+ 48y + 120 = 0

18(x2 - 8x) + 12(y+ 4y) + 120 = 0

18[(x - 4)2 - 16] + 12[(y + 2)2  - 4] + 120 = 0

18(x - 4)- 288 + 12(y + 2)2  - 48 + 120 = 0

18(x - 4)+ 12(y + 2)2  - 336 + 120 = 0

18(x - 4)+ 12(y + 2)2  - 216  =  0

18(x - 4)+ 12(y + 2)2  =  216

[(x - 4)2/12] + [(y + 2)2/18]  =  1

The given conic represents the " Ellipse "

The given ellipse is symmetric about y - axis.

a2  =  18 and b =  12

a  =  3√2 and b  =  2√3

c2  =  a2 - b2

c2  =  18 - 12

c2  =  6

c  =  √6

e = c/a  =  √6/3√2

e = 1/√3

Center :

C (h, k)  ==>  (4, -2)

Foci :

F1 (h, k - c) F2 (h, k + c)

F1 (4, -2 - √6) , F2 (4, -2 + √6) 

Vertices on major axis  :

A (h, k - a) A' (h, k + a)

A (4, -2-3√2) A' (4, -2 + 3√2)

Equation of directrices :

y  =  k ± (a/e)

y  =  -2 ± (3√2/(1/√3))

y  =  -2 ± 3√6

y  =  -2 + 3√6

y  =  -2 - 3√6

(vi) 9x2 − y2 − 36x − 6y + 18 = 0

Solution :

9x2 − y2 − 36x − 6y + 18 = 0

9x2− 36x − y2 − 6y + 18 = 0

9[x2− 4x] − [y2 + 6y] + 18 = 0

9[(x - 2)2− 4] − [(y + 3)2 - 9] + 18 = 0

9(x - 2)2− 36 − (y + 3)2 + 9 + 18 = 0

9(x - 2)- (y + 3)2 = 9

[(x - 2)2/1] - [(y + 3)2/9] = 1

The given conic represents the " Hyperbola "

The given ellipse is symmetric about x - axis.

a2  =  1 and b =  9

a  =  1 and b  =  3

c2  =  a2 + b2

c2  =  1 + 9

c2  =  10

c  =  √10

e = c/a  =  √10/1

e  =  √10

Center :

C (h, k)  ==>  (2, -3)

Foci :

F1 (h + c, k ) F2 (h - c, k )

F1 (2 + √10, -3) F2 (2 - √10, -3)

Vertices on transverse axis  :

A (h + a, k) A' (h - a, k)

A (2 + 1, -3) A' (2 - 1, -3 )

A (3, -3) A' (1, -3)

Equation of directrices :

x  =  h ± (a/e)

x  =  2 ± (1/√10)

x  =  2 + (1/√10)

x  =  2 - (1/√10)

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