This problem asks us to derive the equation of a parabola from its focus-directrix definition. A parabola is the set of all points that are equidistant from a fixed point (the focus) and a fixed line (the directrix). For any point (x,y) on the parabola, the distance to focus F(2,1) equals the distance to directrix x=-2. The distance to F(2,1) is √[(x-2)² + (y-1)²], and the distance to the vertical line x=-2 is |x-(-2)| = |x+2|. Setting these equal and squaring both sides gives (x-2)² + (y-1)² = (x+2)². Expanding and simplifying: x²-4x+4 + (y-1)² = x²+4x+4, which reduces to (y-1)² = 8x. A common error is confusing which variable gets squared based on the directrix orientation. Since we have a vertical directrix and the parabola opens horizontally (to the right), the y-term is squared.

This problem involves deriving a parabola's equation from its focus-directrix definition. A parabola consists of all points equidistant from focus F(0,2) and directrix y=-2. For any point (x,y) on the parabola, we have distance to focus = distance to directrix. The distance to F(0,2) is √[x² + (y-2)²], while the distance to the horizontal line y=-2 is |y+2|. Setting equal and squaring: x² + (y-2)² = (y+2)². Expanding gives x² + y²-4y+4 = y²+4y+4, which simplifies to x² = 8y. This matches the marked answer x² = 8y perfectly. Students might mistakenly include shifts when the vertex is at the origin. The parabola opens upward with vertex at (0,0), equidistant from focus and directrix.

From the equation $$(x - 4)^2 = -8(y + 2)$$, we can identify that this is a downward-opening parabola with vertex (4, -2). Comparing to standard form $$(x - h)^2 = 4p(y - k)$$, we have $$4p = -8$$, so $$p = -2$$. The absolute value $$|p| = 2$$ represents the distance from the vertex to the focus (and also from the vertex to the directrix). The total distance between focus and directrix is $$2|p| = 2(2) = 4$$ units. Choice A gives only |p| instead of 2|p|. Choice C incorrectly calculates as 3|p|. Choice D mistakes the coefficient -8 as the distance.

Since the directrix is vertical (x = -5), this is a horizontal parabola. The vertex is (-1, 3) and directrix is x = -5, so the distance from vertex to directrix is |-1 - (-5)| = 4. Therefore, the focus is 4 units from the vertex in the opposite direction: (-1 + 4, 3) = (3, 3). For a horizontal parabola with vertex (h, k), the equation is $$(y - k)^2 = 4p(x - h)$$ where p is the distance from vertex to focus. Here p = 4, so the equation is $$(y - 3)^2 = 16(x + 1)$$. Choice A uses p = 2 instead of 4. Choice C incorrectly assumes a vertical parabola. Choice D uses 4p = 4 instead of 4p = 16.

With vertex at origin (0, 0) and opening upward, the standard form is $$x^2 = 4py$$ where p is the distance from vertex to focus. Given that the focus is 6 inches from the vertex, $$p = 6$$. Therefore, the equation is $$x^2 = 4(6)y = 24y$$. Choice A uses p directly instead of 4p. Choice B uses 4p/2 = 12 instead of 4p = 24. Choice D incorrectly assumes the parabola opens horizontally instead of vertically, and uses the wrong variable arrangement.

The skill involves deriving the equation of a parabola given its focus and directrix. A parabola is defined as the set of all points equidistant from a fixed point called the focus and a fixed line called the directrix. For any point on the parabola, the distance to the focus F(5,1) equals the distance to the directrix x=1. The vertex is the midpoint between the focus and directrix, calculated as x=(5+1)/2=3 and y=1. This identifies the vertex at (3,1), consistent with the parabola opening to the right. A distractor misconception is miscalculating the midpoint, such as averaging y-coordinates incorrectly leading to (5,3). To derive equations for other parabolas, always start from the distance definition and simplify step by step.

The skill involves deriving the equation of a parabola given its focus and directrix. A parabola is defined as the set of all points equidistant from a fixed point called the focus and a fixed line called the directrix. For any point (x,y) on the parabola, the distance to the focus F(0,5) equals the distance to the directrix y=1. This equality is set as $sqrt(x^2$ + $(y-5)^2$) = |y-1|, which simplifies after squaring to $x^2$ = 8(y-3). The reasoning justifies the upward opening with vertex at (0,3) and 4p=8 from the distance calculations. A distractor misconception is swapping the focus and directrix distances, like setting $sqrt(x^2$ + $(y-1)^2$) = |y-5|, which reverses the roles. To derive equations for other parabolas, always start from the distance definition and simplify step by step.

The skill is deriving the equation of a parabola from its focus and directrix. A parabola is geometrically defined as the set of points equidistant from the focus at (3,-2) and the directrix y = 2. For any point (x,y) on the parabola, the distance to the focus equals the distance to the directrix, expressed as √((x-3)² + (y+2)²) = |y - 2|. Setting up this equality and squaring both sides eliminates the square root, leading to (x-3)² + (y+2)² = (y-2)², which simplifies through expansion and cancellation to (x-3)² = -8y. This final form is justified as it places the vertex at (3,0), midway between the focus and directrix, with the coefficient -8 corresponding to 4p where p=-2 for downward opening. A common distractor misconception is including an unnecessary y-shift like (y+2), altering the vertex position. To transfer this strategy, always start from the distance definition and carefully simplify when deriving parabola equations.

The skill involves deriving the equation of a parabola given its focus and directrix. A parabola is defined as the set of all points equidistant from a fixed point called the focus and a fixed line called the directrix. For any point on the parabola, the distance to the focus F(5,1) equals the distance to the directrix x=1. The vertex is the midpoint between the focus and directrix, calculated as x=(5+1)/2=3 and y=1. This identifies the vertex at (3,1), consistent with the parabola opening to the right. A distractor misconception is miscalculating the midpoint, such as averaging y-coordinates incorrectly leading to (5,3). To derive equations for other parabolas, always start from the distance definition and simplify step by step.

This problem requires deriving a parabola's equation from its focus-directrix definition. A parabola consists of all points equidistant from focus F(2,0) and directrix x=6. For any point (x,y) on the parabola, we equate distances to focus and directrix. The distance to F(2,0) is √[(x-2)² + y²], while the distance to the vertical line x=6 is |x-6|. Setting equal and squaring: (x-2)² + y² = (x-6)². Expanding gives x²-4x+4 + y² = x²-12x+36, which simplifies to y² = -8x+32 = -8(x-4). This matches the marked answer y² = -8(x-4). Students might confuse the sign, but the negative coefficient correctly indicates leftward opening. The vertex is at (4,0), midway between focus and directrix.