The parabola with focus at (4, -3) and vertex | vedclass

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Question

(D) The focus is at $(4, -3)$ and the vertex is at $(4, -1)$.Since the $x$-coordinates are the same,the axis of the parabola is the vertical line $x=4

Vedclass · Accepted Answer

$x^2-8x+8y+24=0$

Vedclass · Answer

(D) The focus is at $(4, -3)$ and the vertex is at $(4, -1)$.Since the $x$-coordinates are the same,the axis of the parabola is the vertical line $x=4$.Since the focus lies below the vertex,the parabola opens downwards.The distance $a$ between the vertex $(4, -1)$ and the focus $(4, -3)$ is $a = |-1 - (-3)| = 2$.The standard equation for a downward-opening parabola with vertex $(h, k)$ is $(x-h)^2 = -4a(y-k)$.Substituting $h=4, k=-1, a=2$:$(x-4)^2 = -4(2)(y - (-1))$$(x-4)^2 = -8(y+1)$$x^2 - 8x + 16 = -8y - 8$$x^2 - 8x + 8y + 24 = 0$.

The parabola with focus at $(4, -3)$ and vertex at $(4, -1)$ is

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