Writing Equations of Parabolas worksheet with 14 problems involving vertex, focus, and directrix.

Let's solve each problem step by step using the vertex form of a parabola. The vertex form depends on the direction the parabola opens.

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🔷 Vertex Form of a Parabola

For a parabola with vertex $(h, k)$:

- If it opens up or down:
$$
(x - h)^2 = 4p(y - k)
$$
- Opens up if $p > 0$, down if $p < 0$

- If it opens left or right:
$$
(y - k)^2 = 4p(x - h)
$$
- Opens right if $p > 0$, left if $p < 0$

Where:
- $p$ = distance from vertex to focus (or from vertex to directrix)
- Focus is at: $(h, k + p)$ for vertical parabolas; $(h + p, k)$ for horizontal
- Directrix: $y = k - p$ for vertical; $x = h - p$ for horizontal

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We'll go through each problem.

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1) Vertex at origin, Focus: $\left(0, -\frac{1}{32}\right)$

- Vertex: $(0, 0)$
- Focus: $(0, -1/32)$ → Below vertex → Opens down
- So: $p = -\frac{1}{32}$
- Vertical parabola → Use: $x^2 = 4p y$
$$
x^2 = 4\left(-\frac{1}{32}\right)y = -\frac{1}{8}y
\Rightarrow \boxed{x^2 = -\frac{1}{8}y}
$$

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2) Vertex at origin, Focus: $\left(0, \frac{1}{8}\right)$

- Vertex: $(0, 0)$
- Focus: $(0, 1/8)$ → Above vertex → Opens up
- $p = \frac{1}{8}$
- Vertical: $x^2 = 4p y$
$$
x^2 = 4\left(\frac{1}{8}\right)y = \frac{1}{2}y
\Rightarrow \boxed{x^2 = \frac{1}{2}y}
$$

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3) Vertex at origin, Directrix: $y = \frac{1}{4}$

- Vertex: $(0, 0)$
- Directrix: $y = 1/4$ → Above vertex → Opens down
- Distance from vertex to directrix: $p = -\frac{1}{4}$ (since it’s below the vertex in direction)
- Wait: Directrix is above, so parabola opens down, and $p = -\frac{1}{4}$
- Use: $x^2 = 4p y$
$$
x^2 = 4(-\frac{1}{4})y = -y
\Rightarrow \boxed{x^2 = -y}
$$

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4) Vertex at origin, Directrix: $y = -\frac{1}{8}$

- Directrix: $y = -1/8$ → Below vertex → Opens up
- So $p = \frac{1}{8}$
- $x^2 = 4p y = 4(\frac{1}{8})y = \frac{1}{2}y$
$$
\boxed{x^2 = \frac{1}{2}y}
$$

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5) Vertex: $(-5, 8)$, Focus: $\left(-\frac{21}{4}, 8\right)$

- Vertex: $(-5, 8)$
- Focus: $(-21/4, 8)$ → Same $y$, so horizontal parabola
- Compute $p$: difference in $x$
$$
p = -\frac{21}{4} - (-5) = -\frac{21}{4} + \frac{20}{4} = -\frac{1}{4}
$$
- So $p = -1/4$ → Opens left
- Use: $(y - k)^2 = 4p(x - h)$
$$
(y - 8)^2 = 4(-\frac{1}{4})(x + 5) = -1(x + 5)
\Rightarrow \boxed{(y - 8)^2 = -(x + 5)}
$$

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6) Vertex: $(-8, -9)$, Focus: $\left(\frac{31}{4}, -9\right)$

- Same $y$ → Horizontal
- $p = \frac{31}{4} - (-8) = \frac{31}{4} + \frac{32}{4} = \frac{63}{4}$
- So $p = 63/4$ → Positive → Opens right
- $(y + 9)^2 = 4p(x + 8)$
$$
(y + 9)^2 = 4 \cdot \frac{63}{4}(x + 8) = 63(x + 8)
\Rightarrow \boxed{(y + 9)^2 = 63(x + 8)}
$$

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7) Vertex: $(-6, -9)$, Directrix: $x = \frac{47}{8}$

- Directrix: $x = 47/8$, vertex at $x = -6 = -48/8$
- So directrix is to the right of vertex → Opens left
- $p = -\left(\frac{47}{8} - (-6)\right) = -\left(\frac{47}{8} + \frac{48}{8}\right) = -\frac{95}{8}$
- Horizontal: $(y + 9)^2 = 4p(x + 6)$
$$
(y + 9)^2 = 4(-\frac{95}{8})(x + 6) = -\frac{380}{8}(x + 6) = -\frac{95}{2}(x + 6)
\Rightarrow \boxed{(y + 9)^2 = -\frac{95}{2}(x + 6)}
$$

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8) Vertex: $(8, 9)$, Directrix: $y = \frac{73}{8}$

- Directrix: $y = 73/8 = 9.125$, vertex: $y = 9 = 72/8$
- Directrix is above vertex → Opens down
- $p = 9 - \frac{73}{8} = \frac{72}{8} - \frac{73}{8} = -\frac{1}{8}$
- Vertical: $(x - 8)^2 = 4p(y - 9)$
$$
(x - 8)^2 = 4(-\frac{1}{8})(y - 9) = -\frac{1}{2}(y - 9)
\Rightarrow \boxed{(x - 8)^2 = -\frac{1}{2}(y - 9)}
$$

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9) Vertex: $(8, -1)$, y-intercept: $-17$

- Vertex: $(8, -1)$
- Y-intercept means $x = 0$, $y = -17$
- Since vertex is not at origin, we use vertex form.
- Assume vertical parabola: $(x - 8)^2 = 4p(y + 1)$
- Plug in point $(0, -17)$:
$$
(0 - 8)^2 = 4p(-17 + 1) \Rightarrow 64 = 4p(-16) = -64p
\Rightarrow p = -1
$$
- So equation:
$$
(x - 8)^2 = 4(-1)(y + 1) = -4(y + 1)
\Rightarrow \boxed{(x - 8)^2 = -4(y + 1)}
$$

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10) Vertex: $(5, -1)$, y-intercept: $-\frac{27}{2}$

- Vertex: $(5, -1)$
- Point: $(0, -27/2)$
- Assume vertical: $(x - 5)^2 = 4p(y + 1)$
- Plug in $x=0$, $y = -27/2$:
$$
(0 - 5)^2 = 4p(-27/2 + 1) = 4p(-25/2)
\Rightarrow 25 = 4p \cdot (-25/2) = -50p
\Rightarrow p = -\frac{1}{2}
$$
- Equation:
$$
(x - 5)^2 = 4(-\frac{1}{2})(y + 1) = -2(y + 1)
\Rightarrow \boxed{(x - 5)^2 = -2(y + 1)}
$$

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11) Opens left or right, Vertex: $(7, 6)$, Passes through: $(-11, 9)$

- Opens left/right → horizontal parabola
- Vertex: $(7, 6)$ → Use: $(y - 6)^2 = 4p(x - 7)$
- Plug in $(-11, 9)$:
$$
(9 - 6)^2 = 4p(-11 - 7) \Rightarrow 9 = 4p(-18) = -72p
\Rightarrow p = -\frac{1}{8}
$$
- So:
$$
(y - 6)^2 = 4(-\frac{1}{8})(x - 7) = -\frac{1}{2}(x - 7)
\Rightarrow \boxed{(y - 6)^2 = -\frac{1}{2}(x - 7)}
$$

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12) Opens left or right, Vertex: $(7, 0)$, Passes through: $(6, -1)$

- Horizontal parabola: $(y - 0)^2 = 4p(x - 7)$ → $y^2 = 4p(x - 7)$
- Plug in $(6, -1)$:
$$
(-1)^2 = 4p(6 - 7) \Rightarrow 1 = 4p(-1) = -4p
\Rightarrow p = -\frac{1}{4}
$$
- So:
$$
y^2 = 4(-\frac{1}{4})(x - 7) = -1(x - 7)
\Rightarrow \boxed{y^2 = -(x - 7)}
$$

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13) Focus: $\left(\frac{63}{8}, -7\right)$, Directrix: $x = \frac{65}{8}$

- Focus: $x = 63/8$, directrix: $x = 65/8$
- Both have same $y = -7$, so horizontal parabola
- Vertex is midway between focus and directrix:
$$
x = \frac{63/8 + 65/8}{2} = \frac{128/8}{2} = \frac{16}{2} = 8
$$
- So vertex: $(8, -7)$
- Distance from vertex to focus: $|63/8 - 8| = |63/8 - 64/8| = 1/8$
- Since focus is left of vertex ($63/8 < 8$), opens left, so $p = -1/8$
- Equation: $(y + 7)^2 = 4p(x - 8)$
$$
(y + 7)^2 = 4(-\frac{1}{8})(x - 8) = -\frac{1}{2}(x - 8)
\Rightarrow \boxed{(y + 7)^2 = -\frac{1}{2}(x - 8)}
$$

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14) Focus: $\left(\frac{107}{12}, -7\right)$, Directrix: $x = \frac{109}{12}$

- Focus: $x = 107/12$, directrix: $x = 109/12$
- Midpoint (vertex): $x = \frac{107/12 + 109/12}{2} = \frac{216/12}{2} = \frac{18}{2} = 9$
- So vertex: $(9, -7)$
- Distance: $|107/12 - 9| = |107/12 - 108/12| = 1/12$
- Focus is left of vertex → $p = -1/12$
- Equation: $(y + 7)^2 = 4p(x - 9)$
$$
(y + 7)^2 = 4(-\frac{1}{12})(x - 9) = -\frac{1}{3}(x - 9)
\Rightarrow \boxed{(y + 7)^2 = -\frac{1}{3}(x - 9)}
$$

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✔ Final Answers:

1. $x^2 = -\frac{1}{8}y$
2. $x^2 = \frac{1}{2}y$
3. $x^2 = -y$
4. $x^2 = \frac{1}{2}y$
5. $(y - 8)^2 = -(x + 5)$
6. $(y + 9)^2 = 63(x + 8)$
7. $(y + 9)^2 = -\frac{95}{2}(x + 6)$
8. $(x - 8)^2 = -\frac{1}{2}(y - 9)$
9. $(x - 8)^2 = -4(y + 1)$
10. $(x - 5)^2 = -2(y + 1)$
11. $(y - 6)^2 = -\frac{1}{2}(x - 7)$
12. $y^2 = -(x - 7)$
13. $(y + 7)^2 = -\frac{1}{2}(x - 8)$
14. $(y + 7)^2 = -\frac{1}{3}(x - 9)$

Let me know if you'd like these rewritten in standard form or graphed!