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.

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.

This problem requires deriving a parabola's equation from its focus-directrix definition. A parabola is the locus of points P(x,y) equidistant from focus F(3,2) and directrix x=7. The distance from P to F is √((x-3)²+(y-2)²), and the distance from P to the vertical line x=7 is |x-7|. Setting these equal: √((x-3)²+(y-2)²) = |x-7|. Since the focus (x=3) is left of the directrix (x=7), the parabola opens leftward, and x < 7 for points on the parabola, so |x-7| = 7-x. Squaring both sides: (x-3)²+(y-2)² = (7-x)², which simplifies to (y-2)² = -8(x-5). A common mistake is assuming the parabola opens rightward without checking the relative positions of focus and directrix.

This problem requires deriving a parabola's equation from its focus-directrix definition. A parabola is the locus of points equidistant from focus F(2,-1) and directrix y=3. For any point P(x,y) on the parabola, we have: distance from P to F equals distance from P to directrix. This gives us √((x-2)²+(y+1)²) = |y-3|. Since the focus is below the directrix, the parabola opens downward, and y < 3 for points on the parabola, so |y-3| = 3-y. Squaring both sides: (x-2)²+(y+1)² = (3-y)², which simplifies to (x-2)² = -8(y-1). A common mistake is assuming the parabola always opens upward or rightward without checking the focus-directrix positions.

To derive this parabola's equation, we apply the focus-directrix definition. A parabola contains all points P(x,y) equidistant from focus F(4,1) and directrix x=0. The distance from P to F is √((x-4)²+(y-1)²), while the distance from P to the vertical line x=0 is |x-0| = |x|. Setting these equal: √((x-4)²+(y-1)²) = |x|. Since the focus is at x=4 (right of the directrix), points on the parabola have x > 0, so |x| = x. Squaring both sides: (x-4)²+(y-1)² = x², which expands and simplifies to (y-1)² = 16(x-2). Students often forget that 4p represents the distance from vertex to focus, leading to incorrect coefficients.

This problem requires deriving a parabola's equation from its focus-directrix definition. A parabola is the locus of points P(x,y) equidistant from focus F(3,2) and directrix x=7. The distance from P to F is √((x-3)²+(y-2)²), and the distance from P to the vertical line x=7 is |x-7|. Setting these equal: √((x-3)²+(y-2)²) = |x-7|. Since the focus (x=3) is left of the directrix (x=7), the parabola opens leftward, and x < 7 for points on the parabola, so |x-7| = 7-x. Squaring both sides: (x-3)²+(y-2)² = (7-x)², which simplifies to (y-2)² = -8(x-5). A common mistake is assuming the parabola opens rightward without checking the relative positions of focus and directrix.

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(2,3) equals the distance to the directrix x=-2. This equality leads to the equation $sqrt((x-2)^2$ + $(y-3)^2$) = |x + 2|, which simplifies after squaring to $(y-3)^2$ = 8x. The final form $(y-3)^2$ = 8(x-0) justifies the vertex at (0,3) and opening to the right with parameter 4p=8 based on the distance from vertex to focus being 2. A distractor misconception is swapping x and y terms, like in choice B, which assumes vertical opening instead of horizontal. 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(-4,0) equals the distance to the directrix x=0. This equality leads to the equation $sqrt((x+4)^2$ + $y^2$) = |x|, which simplifies after squaring to $y^2$ = -8(x+2). The final form $y^2$ = -8(x+2) justifies the vertex at (-2,0) and opening to the left with parameter 4p=8 based on the distance from vertex to focus being 2. A distractor misconception is using a positive coefficient, like in choice B, which would incorrectly open to the right. 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.