The equation of the parabola whose vertex and | vedclass

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(C) Given: Vertex $(h, k) = (0, 4)$ and focus $(0, 2)$.Since the $x$-coordinate of the vertex and focus are the same,the axis of the parabola is verti

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$x^2 + 8y = 32$

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(C) Given: Vertex $(h, k) = (0, 4)$ and focus $(0, 2)$.Since the $x$-coordinate of the vertex and focus are the same,the axis of the parabola is vertical.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 downward-opening parabola is $(x - h)^2 = -4a(y - k)$.Substituting the values: $(x - 0)^2 = -4(2)(y - 4)$.$x^2 = -8(y - 4)$.$x^2 = -8y + 32$.$x^2 + 8y = 32$.

The equation of the parabola whose vertex and focus are $(0, 4)$ and $(0, 2)$ respectively,is

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