Derive the Equation of a Parabola - Geometry

1. A parabola has the equation $(x - 4)^2 = -8(y + 2)$. What is the distance between the focus and the directrix of this parabola?
   Answer: 4 units 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.
2. On the coordinate plane, the focus is $F(-3,1)$ and the directrix is the vertical line $x=1$. The parabola opens to the left. Which equation follows from the focus-directrix definition?
   Answer: $(y-1)^2=-8(x+1)$ 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 (-3,1) and the directrix x = 1. 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-1)²) = |x - 1|. Setting up this equality and squaring both sides eliminates the square root, leading to (x+3)² + (y-1)² = (x-1)², which simplifies through expansion and cancellation to (y-1)² = -8(x+1). This final form is justified as it places the vertex at (-1,1), midway between the focus and directrix, with the coefficient -8 corresponding to 4p where p=-2 for leftward opening. A common distractor misconception is using a positive coefficient, implying incorrect opening direction. To transfer this strategy, always start from the distance definition and carefully simplify when deriving parabola equations.
3. A parabola has focus $F(0,-4)$ and directrix $y=2$. Which statement identifies the vertex?
   Answer: The vertex is at $(0,-1)$. This problem asks us to find the vertex of a parabola given its focus and directrix. A parabola is defined as the set of points equidistant from focus F(0,-4) and directrix y=2. The vertex is the point on the parabola closest to both the focus and directrix, located exactly halfway between them. Since the focus is at y=-4 and the directrix is at y=2, the vertex's y-coordinate is (-4+2)/2 = -1. The vertex shares the same x-coordinate as the focus for vertical parabolas, so the vertex is at (0,-1). A common error is placing the vertex at the focus or confusing the vertex with other special points. Always remember: the vertex lies on the axis of symmetry, midway between focus and directrix.
4. A parabola is the set of points equidistant from the focus $F(-2,0)$ and the directrix $x=2$. The parabola opens to the left. Which expression represents all points equidistant from the focus and directrix?
   Answer: $\sqrt{(x+2)^2+y^2}=|x-2|$ 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 (-2,0) and the directrix x = 2. For any point (x,y) on the parabola, the distance to the focus equals the distance to the directrix, expressed directly as √((x+2)² + y²) = |x - 2| before squaring. This expression arises by applying the distance formulas without simplification, capturing the equality condition. The form is justified as it correctly uses absolute value for the directrix distance and square root for the point-to-focus distance. A common distractor misconception is omitting the square root or using incorrect coordinates like |y-2|. To transfer this strategy, always start from the distance definition and set up the unsimplified equality when deriving parabola equations.
5. A parabola is the set of points equidistant from focus $F(0,2)$ and the directrix $y=-2$ (opening upward). Which equation represents the parabola?
   Answer: $x^2=8y$ 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.
6. A parabola is the set of all points equidistant from the focus $F(0,2)$ and the directrix $y=-2$. The parabola opens upward. Which reasoning correctly derives the equation?
   Answer: Set $\sqrt{x^2+(y-2)^2}=|y+2|$, square, and simplify to $x^2=8y$. 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 (0,2) and the directrix y = -2. For any point (x,y) on the parabola, the distance to the focus equals the distance to the directrix, correctly expressed as √(x² + (y-2)²) = |y + 2|. Setting up this equality and squaring both sides eliminates the square root, leading to x² + (y-2)² = (y+2)², which simplifies through expansion and cancellation to x² = 8y. This final form is justified as it places the vertex at (0,0), midway between the focus and directrix, with the coefficient 8 corresponding to 4p where p=2 for upward opening. A common distractor misconception is swapping the signs in the distance expression, like using |y-2| for the directrix. To transfer this strategy, always start from the distance definition and carefully simplify when deriving parabola equations.
7. The equation of a parabola can be written as $y = ax^2 + bx + c$. If this parabola has focus at $(1, 3)$ and vertex at $(1, 2)$, what is the value of coefficient $a$?
   Answer: $a = \frac{1}{4}$ When you encounter a parabola problem involving the focus and vertex, you're working with the geometric definition of a parabola. The key relationship is that the distance from the vertex to the focus equals $\frac{1}{4|a|}$, where $a$ is the coefficient of $x^2$. Since both the vertex $(1, 2)$ and focus $(1, 3)$ have the same x-coordinate, this parabola opens vertically. The distance from vertex to focus is $|3 - 2| = 1$. Using the relationship $\frac{1}{4|a|} = 1$, we can solve: $4|a| = 1$, so $|a| = \frac{1}{4}$. Since the focus is above the vertex, the parabola opens upward, making $a$ positive. Therefore, $a = \frac{1}{4}$. Looking at the wrong answers: Choice A gives $a = \frac{1}{2}$, which would place the focus at distance $\frac{1}{4 \cdot \frac{1}{2}} = \frac{1}{2}$ from the vertex, not 1 unit away. Choice B suggests $a = 4$, making the focus-vertex distance $\frac{1}{16}$, far too small. Choice C offers $a = \frac{1}{8}$, which would put the focus 2 units from the vertex instead of 1. Remember this pattern: for a parabola in standard form, the focus-vertex distance is always $\frac{1}{4|a|}$. When you know this distance from the coordinates, set up the equation $\frac{1}{4|a|} = \text{distance}$ and solve for $a$. The sign of $a$ depends on whether the parabola opens up (positive) or down (negative).
8. A parabola has focus $F(3, -2)$ and directrix $y = 4$. If point $P(x, y)$ lies on this parabola, which equation represents the relationship that $P$ must satisfy?
   Answer: $(x - 3)^2 = -12(y - 1)$ For a point P(x,y) on a parabola, the distance from P to the focus equals the distance from P to the directrix. Distance from P to F(3,-2): $\sqrt{(x-3)^2 + (y+2)^2}$. Distance from P to directrix y=4: $|y-4|$. Setting these equal and squaring: $(x-3)^2 + (y+2)^2 = (y-4)^2$. Expanding: $(x-3)^2 + y^2 + 4y + 4 = y^2 - 8y + 16$. Simplifying: $(x-3)^2 = -12y + 12 = -12(y-1)$. Choice B uses incorrect coefficient -6 instead of -12. Choice C has wrong sign (positive instead of negative). Choice D incorrectly treats this as a horizontal parabola.
9. A parabola has directrix $y = x + 4$ and focus at $F(2, 0)$. Using the definition of a parabola, which equation must be satisfied by any point $(x, y)$ on this parabola?
   Answer: $\sqrt{(x-2)^2 + y^2} = \frac{|x - y + 4|}{\sqrt{2}}$ When you encounter a parabola problem with a given focus and directrix, remember that a parabola is defined as the set of all points equidistant from the focus and the directrix line. For any point $(x, y)$ on the parabola, you need to set up two distance calculations. The distance from point $(x, y)$ to the focus $F(2, 0)$ uses the standard distance formula: $\sqrt{(x-2)^2 + y^2}$. The distance from point $(x, y)$ to the line $y = x + 4$ (or $x - y + 4 = 0$) requires the point-to-line distance formula: $\frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}}$. Here, $a = 1$, $b = -1$, $c = 4$, so the distance is $\frac{|x - y + 4|}{\sqrt{1^2 + (-1)^2}} = \frac{|x - y + 4|}{\sqrt{2}}$. Setting these distances equal gives us answer choice D. Choice A incorrectly squares both sides without using the proper point-to-line formula. Choice B uses $|x - y + 4|$ as the distance to the line, but this is the numerator of the distance formula—it's missing the $\sqrt{2}$ denominator. Choice C has the correct denominator but incorrectly writes the numerator as $|y - x - 4|$ instead of $|x - y + 4|$, which changes the sign and creates an entirely different line. Study tip: Always use the point-to-line distance formula $\frac{|ax + by + c|}{\sqrt{a^2 + b^2}}$ when finding distance from a point to a line—don't forget that denominator!
10. A parabola opens upward with its focus at $(2, 7)$ and vertex at $(2, 4)$. If the parabola passes through point $(6, y)$, what is the value of $y$?
    Answer: $y = \frac{20}{3}$ First, find the parabola equation. The vertex is (2, 4) and focus is (2, 7), so p = 7 - 4 = 3. The equation is $(x - 2)^2 = 4p(y - 4) = 12(y - 4)$. For point (6, y): $(6 - 2)^2 = 12(y - 4)$, so $16 = 12(y - 4)$. Solving: $\frac{16}{12} = y - 4$, so $\frac{4}{3} = y - 4$, giving $y = 4 + \frac{4}{3} = \frac{12 + 4}{3} = \frac{20}{3}$. Choice A incorrectly assumes y equals the focus y-coordinate. Choice B results from arithmetic error: $\frac{16}{12} + 4 = \frac{4}{3} + \frac{12}{3} = \frac{16}{3}$ but written as $\frac{19}{3}$. Choice D uses incorrect formula or calculation.
11. A parabola is defined by focus $F(-2,-3)$ and directrix $y=-1$. Which equation follows from the focus-directrix definition?​
    Answer: $(x+2)^2=-2(y+2)$ To derive this parabola's equation, we use the focus-directrix definition. A parabola consists of all points P(x,y) equidistant from focus F(-2,-3) and directrix y=-1. The distance from P to F is √((x+2)²+(y+3)²), while the distance from P to the horizontal line y=-1 is |y-(-1)| = |y+1|. Setting these equal: √((x+2)²+(y+3)²) = |y+1|. Since the focus (y=-3) is below the directrix (y=-1), the parabola opens downward, and y < -1 for points on the parabola, so |y+1| = -1-y. Squaring both sides: (x+2)²+(y+3)² = (-1-y)², which simplifies to (x+2)² = -2(y+2). Students often forget to determine the parabola's orientation before removing absolute values.
12. A parabola is defined as the set of all points equidistant from a focus and a directrix. The focus is $F(0,2)$ and the directrix is the line $y=-2$. Which equation follows from the focus–directrix definition?
    Answer: $x^2=8y$ 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 P(x,y) on the parabola, the distance from P to focus F(0,2) equals the distance from P to the directrix y=-2. Setting these distances equal: √(x²+(y-2)²) = |y-(-2)| = |y+2|. Since y > -2 for points above the directrix, we have |y+2| = y+2, so √(x²+(y-2)²) = y+2. Squaring both sides and simplifying: x²+(y-2)² = (y+2)², which gives x² = 8y. A common error is forgetting to square both sides properly or misidentifying which direction the parabola opens.
13. Let $P(x,y)$ be a point on a parabola with focus $F(1,0)$ and directrix $x=-1$. Which expression represents all points equidistant from the focus and directrix?​
    Answer: $\sqrt{(x-1)^2+y^2}=|x+1|$ This problem asks for the distance equation that defines a parabola. A parabola is the set of all points P(x,y) equidistant from focus F(1,0) and directrix x=-1. The distance from P to the focus F(1,0) is √((x-1)²+(y-0)²) = √((x-1)²+y²). The distance from P to the vertical line x=-1 is the horizontal distance |x-(-1)| = |x+1|. By the focus-directrix definition, these distances are equal: √((x-1)²+y²) = |x+1|. This equation represents all points on the parabola before algebraic simplification. A common error is dropping the square root on the focus distance or using the wrong distance formula for the directrix. Always start with the geometric definition before simplifying.
14. On the coordinate plane, the focus is $F(5,1)$ and the directrix is the vertical line $x=1$ (shown). The parabola opens to the right. Which statement identifies the vertex?​
    Answer: The vertex is at $(3,1)$. 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.
15. A parabola is defined by focus $F(0,5)$ and directrix $y=1$ (shown on the coordinate plane). The parabola opens upward. Which reasoning correctly derives the equation?​
    Answer: Set $\sqrt{x^2+(y-5)^2}=|y-1|$ and simplify to $x^2=8(y-3)$. 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.
16. Let $P(x,y)$ be a point on a parabola with focus $F(0,5)$ and directrix $y=1$. Which expression represents all points equidistant from the focus and directrix?
    Answer: $\sqrt{x^2+(y-5)^2}=|y-1|$ To write the distance equation for this parabola, we apply the focus-directrix definition. A parabola contains all points P(x,y) equidistant from focus F(0,5) and directrix y=1. The distance from P to focus F(0,5) is √((x-0)²+(y-5)²) = √(x²+(y-5)²). The distance from P to the horizontal line y=1 is the vertical distance |y-1|. Setting these distances equal gives the defining equation: √(x²+(y-5)²) = |y-1|. This represents the geometric condition before any algebraic manipulation. Students often confuse vertical and horizontal distances or forget the absolute value for the directrix distance. Remember: distance to a horizontal line is |y-k|, while distance to a vertical line is |x-h|.
17. A parabola is defined by focus $F(2,-1)$ and directrix $y=3$. Which equation represents the parabola?
    Answer: $(x-2)^2=-8(y-1)$ 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.
18. A parabola is defined as the set of points equidistant from the focus $F(0,1)$ and the directrix line $y=-3$. The parabola opens upward. Which expression represents all points $(x,y)$ equidistant from the focus and directrix?
    Answer: $\sqrt{x^2+(y-1)^2}=|y+3|$ The skill here is deriving the equation of a parabola from 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 this parabola, the distance to the focus F(0,1) equals the distance to the directrix y=-3. This equality is directly represented by the expression √(x² + (y-1)²) = |y+3|, without yet squaring. This form is justified as it captures the raw geometric definition before algebraic simplification to the parabola equation. A distractor like choice B swaps the focus and directrix distances, which would not satisfy the definition. To derive equations for other parabolas, always start by equating the distance to the focus and to the directrix, then square and simplify if needed.
19. On the coordinate plane, the focus is $F(0,4)$ and the directrix is the horizontal line $y=0$. The parabola opens upward. Which statement identifies the vertex?
    Answer: The vertex is $(0,2)$. 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 (0,4) and the directrix y = 0. For any point (x,y) on the parabola, the distance to the focus equals the distance to the directrix, with the vertex located midway between them. The vertex arises at the midpoint, calculated as (0, (4+0)/2) = (0,2), since the focus and directrix are horizontally aligned in y. This position is justified because the parabola opens upward from the midpoint, equidistant at 2 units from both focus and directrix. A common distractor misconception is placing the vertex at the focus (0,4) or origin (0,0), ignoring the midpoint calculation. To transfer this strategy, always start from the distance definition and identify the midpoint for the vertex when deriving parabola equations.
20. A parabola is the set of all points equidistant from the focus $F(-3,1)$ and the directrix line $x=1$. The parabola opens left. Which statement identifies the vertex?
    Answer: The vertex is at $(-1,1)$. The skill here is deriving properties of a parabola, such as its vertex, from the 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. The vertex lies on the axis of symmetry, equidistant from the focus and directrix, so for focus F(-3,1) and directrix x=1, the distances must be equal at the midpoint. The axis is horizontal, and the vertex x-coordinate is the average of -3 and 1, giving x=-1, with y=1, so (-1,1). This is justified as the perpendicular distance from vertex to directrix equals the distance to focus, both 2 units. A distractor like choice B might take the focus coordinates directly as the vertex, neglecting the midpoint. To find vertices for other parabolas, always start by finding the midpoint between the focus and its perpendicular foot on the directrix.

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