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Question
Find the centre, eccentricity, foci and directrice of the hyperbola.
x2 − y2 + 4x = 0
Solution
Given:
The equation ⇒ x2 – y2 + 4x = 0
Let us find the centre, eccentricity, foci and directions of the hyperbola
By using the given equation
x2 – y2 + 4x = 0
x2 + 4x + 4 – y2 – 4 = 0
(x + 2)2 – y2 = 4
`(x + 2)^2/4 - y^2/4 = 1`
`(x + 2)^2/2^2 - y^2/2^2 = 1`
Here, center of the hyperbola is (2, 0)
So, let x – 2 = X
The obtained equation is of the form
`x^2/a^2 - y^2/b^2 = 1`
Where, a = 2 and b = 2
Eccentricity is given by:
`e = sqrt(1 + b^2/a^2)`
= `sqrt(1 + 4/4)`
= `sqrt(1 + 1)`
= `sqrt2`
Foci: The coordinates of the foci are (± ae, 0)
X = ± 2√2 and Y = 0
X + 2 = ± 2√2 and Y = 0
X= ± 2√2 – 2 and Y = 0
So, Foci = (± 2√2 – 2, 0)
Equation of directrix are:
`X = pm a/e`
⇒ `X = pm 2/sqrt2`
⇒ `X = pm 2/sqrt2`
⇒ `X = pm sqrt2`
⇒ `X mp sqrt2 = 0`
⇒ `x + 2 mp sqrt2 = 0`
x + 2 − √2 = 0 and x + 2 + √2 = 0
∴ The center is (−2, 0), eccentricity (e) = √2, Foci = (−2 ± 2√2, 0), Equation of directrix = x + 2 = ±√2
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