Find the vertex , focus, axis, directrix and latus rectum of the following parabola: `y^2-4y-3x+1=0`

Explanation

`y^2 - 4y = 3x - 1`

`y^2 - 4y + 4 = 3x + 3`

`(y - 2)^2 = 3(x + 1)`

Comparing with the standard form of parabola `(y - a)^2 = 4b(x - c)` we get,

`4b = 3 ⇒ b = frac{3}{ 4} `

The vertex is `(c, a) = (- 1, 2)`

The focus is `(b + c, a) = ( frac{3} {4}-1, 2) = (frac{ − 1} {4},2)`

The equation of the axis is `y - a = 0`.
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