The focus of the parabola 4y^2 - 6x - 4y = 5 | vedclass

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(B) Given equation of the parabola: $4y^2 - 4y - 6x = 5$Completing the square for $y$: $4(y^2 - y) = 6x + 5$$4(y^2 - y + 1/4) = 6x + 5 + 1$$4(y - 1/2)

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$(-5/8, 1/2)$

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(B) Given equation of the parabola: $4y^2 - 4y - 6x = 5$Completing the square for $y$: $4(y^2 - y) = 6x + 5$$4(y^2 - y + 1/4) = 6x + 5 + 1$$4(y - 1/2)^2 = 6(x + 1)$$(y - 1/2)^2 = \frac{3}{2}(x + 1)$Comparing with the standard form $Y^2 = 4aX$,where $Y = y - 1/2$,$X = x + 1$,and $4a = 3/2$.Thus,$a = 3/8$.The focus in the $(X, Y)$ coordinate system is $(a, 0) = (3/8, 0)$.Converting back to $(x, y)$ coordinates:$x + 1 = 3/8 \Rightarrow x = -5/8$$y - 1/2 = 0 \Rightarrow y = 1/2$Therefore,the focus is $(-5/8, 1/2)$.

$A. \ 2x-5=0$	$I. \ \text{Vertex}$
$B. \ (\frac{3}{2}, -1)$	$II. \ \text{Focus}$
$C. \ y+1=0$	$III. \ \text{Equation of directrix}$
$D. \ (2, -1)$	$IV. \ \text{Equation of the axis}$
	$V. \ \text{Equation of the Latus rectum}$

The focus of the parabola $4y^2 - 6x - 4y = 5$ is

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