# Find directrix and axis from this equation of parabola 4y^2- 6x-4y=5

The given parabola

$4y^{2} -6x-4y = 5 \\ \Rightarrow 4(y^{2} - y) = 6x + 4 \\ \Rightarrow (y^{2} - y) = \frac{1}{4}(6x +4)\\ \Rightarrow y^{2} - y +\frac{1}{4} = \frac{1}{4}(6x +5) +\frac{1}{4}\\ \Rightarrow (y-\frac{1}{2})^{2} = \frac{1}{4} (6x +6)\\ \Rightarrow (y-\frac{1}{2})^{2} = \frac{3}{2} (x +1)\\$

we know that the general equation of a parabola

$y^{2} = 4 a x$

Here

$y = y-\frac{1}{2} and, a = \frac{3}{8} and, x = x+ 1$

Directrix  $x = - a$

$x +1 = - \frac{3}{8}\\ 8x + 11 = 0\\$

The axis

$y = 0 \\ y-\frac{1}{2} = 0\\ 2y -1 = 0 \\$

Hence , Directrics

$8x + 11 \\ and Axis = 2y-1 = 0$

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