Question

Find the coordinates of the centre, foci and equation of directrix of the hyperbola x² – 3y² – 4x = 8.

Answer 1

Here, the equation of the given hyporbola is

x² - 3y² - 4x = 8 

⇒ x² - 4x + 4 - 3y² = 8 + 4 

⇒ (x - 2)² - 3y² = 12

⇒ `("x" - 2)^2/12 - "y"^2/4 = 1`

Here, a² = 12 and b² = 4 ⇒ a = `2sqrt3` and b = 2

For centre put X = 0   and Y = 0

x - 2 = 0   and y = 0

x = 2   and   y = 0

∴ Coordinates of the centre are (2,0)

For foci, X = ± ae and Y = 0

Now, e = `sqrt("a"^2 + "b"^2)/"a"^2 = sqrt(12 + 4)/12 = sqrt(16/12) = 4/2sqrt3 = 2/sqrt3`

Answer 2

Here, the equation of the given hyporbola is

x² - 3y² - 4x = 8 

⇒ x² - 4x + 4 - 3y² = 8 + 4 

⇒ (x - 2)² - 3y² = 12

⇒ `("x" - 2)^2/12 - "y"^2/4 = 1`

Writing x - 2 = X and y = Y, the given equation becomes

`"X"^2/12 - "Y"^2/4 = 1`

Here, a² = 12 and b² = 4 ⇒ a = `2sqrt3` and b = 2

For centre put X = 0   and Y = 0

x - 2 = 0   and y = 0

x = 2   and   y = 0

∴ Coordinates of the centre are (2,0)

For foci, X = ± ae and Y = 0

Now, e = `sqrt("a"^2 + "b"^2)/"a"^2 = sqrt(12 + 4)/12 = sqrt(16/12) = 4/(2sqrt3) = 2/sqrt3`

∴ x - 2 = ± `2sqrt3 xx 2/sqrt3 = +-4`

x = ± 4 + 2

⇒ x = 6 , x = -2 and y = 0

∴ Coordinates of Foci are (6,0) and (-2,0) 

The equation of directrices are:

X = `+- "a"/"e"`

x - 2 = `+- (2sqrt3)/(2/sqrt3)`

x - 2 = ± 3

x = ± 3 + 2 ⇒ x = 5 and x = -1

∴ x = 5 and x = -1 are the equations of directrices.