FIND VERTEX FOCUS EQUATION OF DIRECTRIX OF HYPERBOLA

Note :

e  =  √[1 + (b²/a²)]

b²  =  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 b²  =  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)

(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 b²  =  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)

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

Solution :

18x² + 12y² − 144x + 48y + 120 = 0

18x² − 144x + 12y² + 48y + 120 = 0

18(x² - 8x) + 12(y² + 4y) + 120 = 0

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

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

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

18(x - 4)² + 12(y + 2)²  - 216  =  0

18(x - 4)² + 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 b²  =  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

(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)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 b²  =  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)

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