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प्रश्न
Find the coordinates of the centre, foci and equation of directrix of the hyperbola x2 – 3y2 – 4x = 8.
उत्तर १
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`
उत्तर २
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.
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