# Find the Coordinates of the Centre, Foci and Equation of Directrix of the Hyperbola X2 – 3y2 – 4x = 8. - Mathematics

Sum

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

#### Solution 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

#### Solution 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.

Concept: Basic Concepts of Differential Equation
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