The vertex of a parabola is (2, 2) and the co | vedclass

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(C) The vertex of the parabola is $(h, k) = (2, 2)$.Since the extremities of the latus rectum are $(-2, 0)$ and $(6, 0)$,the axis of the parabola is t

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$x^2 - 4x + 8y - 12 = 0$

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(C) The vertex of the parabola is $(h, k) = (2, 2)$.Since the extremities of the latus rectum are $(-2, 0)$ and $(6, 0)$,the axis of the parabola is the vertical line $x = 2$.The parabola opens downwards,so its equation is of the form $(x - h)^2 = -4a(y - k)$.Substituting the vertex $(2, 2)$,we get $(x - 2)^2 = -4a(y - 2)$.The length of the latus rectum is the distance between $(-2, 0)$ and $(6, 0)$,which is $|6 - (-2)| = 8$.Thus,$4a = 8$,so $a = 2$.The equation becomes $(x - 2)^2 = -8(y - 2)$.Expanding this,we get $x^2 - 4x + 4 = -8y + 16$,which simplifies to $x^2 - 4x + 8y - 12 = 0$.

The vertex of a parabola is $(2, 2)$ and the coordinates of the two extremities of its latus rectum are $(-2, 0)$ and $(6, 0)$. The equation of the parabola is

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