If (0, 4) and (0, 2) are the vertex and focus | vedclass

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(C) The vertex is $V(0, 4)$ and the focus is $S(0, 2)$.Since both points lie on the $y$-axis,the axis of the parabola is the $y$-axis.The distance bet

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$x^2 + 8y = 32$

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(C) The vertex is $V(0, 4)$ and the focus is $S(0, 2)$.Since both points lie on the $y$-axis,the axis of the parabola is the $y$-axis.The distance between the vertex and the focus is $a = |4 - 2| = 2$.Since the focus is below the vertex,the parabola opens downwards.The standard equation for a parabola with vertex $(h, k)$ opening downwards is $(x - h)^2 = -4a(y - k)$.Substituting $h = 0, k = 4$,and $a = 2$,we get:$(x - 0)^2 = -4(2)(y - 4)$$x^2 = -8(y - 4)$$x^2 = -8y + 32$$x^2 + 8y = 32$.

If $(0, 4)$ and $(0, 2)$ are the vertex and focus of a parabola respectively,then what is its equation?

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