What are the directrix and axis of the parabo | vedclass

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(C) Given equation: $4y^2 - 4y = 6x + 5$Completing the square for $y$: $4(y^2 - y) = 6x + 5$$4(y^2 - y + \frac{1}{4}) = 6x + 5 + 1$$4(y - \frac{1}{2})

Vedclass · Accepted Answer

$8x + 11 = 0, 2y - 1 = 0$

Vedclass · Answer

(C) Given equation: $4y^2 - 4y = 6x + 5$Completing the square for $y$: $4(y^2 - y) = 6x + 5$$4(y^2 - y + \frac{1}{4}) = 6x + 5 + 1$$4(y - \frac{1}{2})^2 = 6(x + 1)$$(y - \frac{1}{2})^2 = \frac{6}{4}(x + 1) = \frac{3}{2}(x + 1)$Comparing with $(Y)^2 = 4aX$,where $Y = y - \frac{1}{2}$,$X = x + 1$,and $4a = \frac{3}{2} \Rightarrow a = \frac{3}{8}$.The axis is $Y = 0$ $\Rightarrow y - \frac{1}{2} = 0$ $\Rightarrow 2y - 1 = 0$.The directrix is $X + a = 0$ $\Rightarrow x + 1 + \frac{3}{8} = 0$ $\Rightarrow x + \frac{11}{8} = 0$ $\Rightarrow 8x + 11 = 0$.

What are the directrix and axis of the parabola $4y^2 - 6x - 4y = 5$ respectively?

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