Question

9-4 additional practice
parabolas in the coordinate plane
find the vertex of each parabola

1. focus: (2, 9); directrix: y = 1 2. focus: (2, −2); directrix: y = 0

find the vertex, focus, and directrix of each parabola.

1. $y - 4 = \frac{1}{16}(x - 7)^2$ 4. $y + 7 = \frac{1}{12}(x - 6)^2$
2. $y - 3 = \frac{1}{24}(x + 3)^2$ 6. $y - 5 = \frac{1}{2}(x + 3)^2$

write an equation of each parabola with the given focus and vertex.

1. focus: (−4, 9); vertex: (−4, 5) 8. focus: (2, 4); vertex: (2, 0)
2. focus: (−7, 10); vertex: (−7, 4) 10. focus: (−5, −3); vertex: (−5, −6)

Explanation:

Step1: Recall the vertex formula for parabolas

For a parabola in the form \((x - h)^2 = 4p(y - k)\), the vertex is \((h, k)\), focus is \((h, k + p)\), and directrix is \(y = k - p\). Also, the vertex is the midpoint between the focus and the directrix (for vertical parabolas).

Step2: Solve Problem 1 (focus: \((2, 9)\); directrix: \(y = 1\))

The vertex \((h, k)\) is the midpoint of the focus \((h, y_f)\) and the directrix \(y = y_d\). So \(k=\frac{y_f + y_d}{2}=\frac{9 + 1}{2}=5\), and \(h = 2\) (since the focus and directrix have the same \(x\)-coordinate, the parabola is vertical). So vertex is \((2, 5)\).

Step3: Solve Problem 2 (focus: \((2, -2)\); directrix: \(y = 0\))

Using midpoint formula: \(k=\frac{-2 + 0}{2}=-1\), \(h = 2\). Vertex is \((2, -1)\).

Step4: Solve Problem 3 (\(y - 4=\frac{1}{16}(x - 7)^2\))

Rewrite as \((x - 7)^2 = 16(y - 4)\), which is \((x - h)^2 = 4p(y - k)\) with \(h = 7\), \(k = 4\), \(4p = 16\Rightarrow p = 4\). Vertex is \((7, 4)\). Focus: \((7, 4 + 4)=(7, 8)\). Directrix: \(y = 4 - 4 = 0\).

Step5: Solve Problem 4 (\(y + 7=\frac{1}{12}(x - 6)^2\))

Rewrite as \((x - 6)^2 = 12(y + 7)\), so \(h = 6\), \(k = -7\), \(4p = 12\Rightarrow p = 3\). Vertex: \((6, -7)\). Focus: \((6, -7 + 3)=(6, -4)\). Directrix: \(y=-7 - 3=-10\).

Step6: Solve Problem 5 (\(y - 3=\frac{1}{24}(x + 3)^2\))

Rewrite as \((x + 3)^2 = 24(y - 3)\), so \(h=-3\), \(k = 3\), \(4p = 24\Rightarrow p = 6\). Vertex: \((-3, 3)\). Focus: \((-3, 3 + 6)=(-3, 9)\). Directrix: \(y = 3 - 6=-3\).

Step7: Solve Problem 6 (\(y - 5=\frac{1}{2}(x + 3)^2\))

Rewrite as \((x + 3)^2 = 2(y - 5)\), so \(h=-3\), \(k = 5\), \(4p = 2\Rightarrow p=\frac{1}{2}\). Vertex: \((-3, 5)\). Focus: \((-3, 5+\frac{1}{2})=(-3, \frac{11}{2})\). Directrix: \(y = 5-\frac{1}{2}=\frac{9}{2}\).

Step8: Solve Problem 7 (focus: \((-4, 9)\); vertex: \((-4, 5)\))

Vertex \((h, k)=(-4, 5)\), focus \((h, k + p)=(-4, 9)\). So \(k + p = 9\Rightarrow 5 + p = 9\Rightarrow p = 4\). The equation is \((x + 4)^2 = 4p(y - 5)\Rightarrow(x + 4)^2 = 16(y - 5)\).

Step9: Solve Problem 8 (focus: \((2, 4)\); vertex: \((2, 0)\))

Vertex \((h, k)=(2, 0)\), focus \((h, k + p)=(2, 4)\). So \(0 + p = 4\Rightarrow p = 4\). Equation: \((x - 2)^2 = 4p(y - 0)\Rightarrow(x - 2)^2 = 16y\).

Step10: Solve Problem 9 (focus: \((-7, 10)\); vertex: \((-7, 4)\))

Vertex \((h, k)=(-7, 4)\), focus \((h, k + p)=(-7, 10)\). So \(4 + p = 10\Rightarrow p = 6\). Equation: \((x + 7)^2 = 4p(y - 4)\Rightarrow(x + 7)^2 = 24(y - 4)\).

Step11: Solve Problem 10 (focus: \((-5, -3)\); vertex: \((-5, -6)\))

Vertex \((h, k)=(-5, -6)\), focus \((h, k + p)=(-5, -3)\). So \(-6 + p = -3\Rightarrow p = 3\). Equation: \((x + 5)^2 = 4p(y + 6)\Rightarrow(x + 5)^2 = 12(y + 6)\).

Answer:

1. Vertex: \(\boldsymbol{(2, 5)}\)
2. Vertex: \(\boldsymbol{(2, -1)}\)
3. Vertex: \(\boldsymbol{(7, 4)}\), Focus: \(\boldsymbol{(7, 8)}\), Directrix: \(\boldsymbol{y = 0}\)
4. Vertex: \(\boldsymbol{(6, -7)}\), Focus: \(\boldsymbol{(6, -4)}\), Directrix: \(\boldsymbol{y = -10}\)
5. Vertex: \(\boldsymbol{(-3, 3)}\), Focus: \(\boldsymbol{(-3, 9)}\), Directrix: \(\boldsymbol{y = -3}\)
6. Vertex: \(\boldsymbol{(-3, 5)}\), Focus: \(\boldsymbol{(-3, \frac{11}{2})}\), Directrix: \(\boldsymbol{y = \frac{9}{2}}\)
7. Equation: \(\boldsymbol{(x + 4)^2 = 16(y - 5)}\)
8. Equation: \(\boldsymbol{(x - 2)^2 = 16y}\)
9. Equation: \(\boldsymbol{(x + 7)^2 = 24(y - 4)}\)
10. Equation: \(\boldsymbol{(x + 5)^2 = 12(y + 6)}\)