If (0, 6) and (0, 3) are respectively the ver | vedclass

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(A) Given: Vertex $A = (0, 6)$ and Focus $S = (0, 3)$.Since the vertex and focus lie on the $y$-axis,the axis of the parabola is the $y$-axis.The dist

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$x^2 + 12y = 72$

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(A) Given: Vertex $A = (0, 6)$ and Focus $S = (0, 3)$.Since the vertex and focus lie on the $y$-axis,the axis of the parabola is the $y$-axis.The distance $a$ between the vertex and the focus is $a = 6 - 3 = 3$.Since the focus is below the vertex,the parabola opens downwards.The standard form of a downward-opening parabola with vertex $(h, k)$ is $(x - h)^2 = -4a(y - k)$.Here,$(h, k) = (0, 6)$ and $a = 3$.Substituting these values,we get $(x - 0)^2 = -4(3)(y - 6)$.$x^2 = -12(y - 6)$.$x^2 = -12y + 72$.$x^2 + 12y = 72$.

If $(0, 6)$ and $(0, 3)$ are respectively the vertex and focus of a parabola,then its equation is

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