Find the centre, eccentricity, foci and directrices of the hyperbola : `x^2-3y^2-2x=8.`

Explanation

We Have given `x^2 – 3y^2–2x=8`

And to find the center, eccentricity(e), coordinates of the foci f(m, n), equation of directrix. `x^2 – 3y^2 – 2x = 8`
`implies x^2 – 2x + 1 – 3y^2 – 1 = 8`

`implies (x – 1)^2 – 3y^2 = 9`

`implies ((x – 1)^2)/9 – (3y^2)/9 = 1`

`implies ((x – 1)^2)/(3^2) – (y^2)/(sqrt(3)^2) = 1`

As the center of the hyperbola is `(1, 0)`

Then let `x –1=X`

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