Parabolas (Conic Form)

Parabolas with horizontal and vertical orientations

Parabolas (Conic Section)

Vertical Parabola (opens up/down)

(x−h)2=4p(y−k)(x - h)^2 = 4p(y - k)

  • Vertex: (h,k)(h, k)
  • Focus: (h,k+p)(h, k + p)
  • Directrix: y=k−py = k - p
  • Opens up if p>0p > 0, down if p<0p < 0

Horizontal Parabola (opens left/right)

(y−k)2=4p(x−h)(y - k)^2 = 4p(x - h)

  • Vertex: (h,k)(h, k)
  • Focus: (h+p,k)(h + p, k)
  • Directrix: x=h−px = h - p
  • Opens right if p>0p > 0, left if p<0p < 0

What is pp?

pp is the focal distance (distance from vertex to focus)

Converting Forms

From: y=ax2+bx+cy = ax^2 + bx + c

To conic form:

  1. Complete the square
  2. Identify 4p4p from coefficient

Example: y=x2+4x+1y = x^2 + 4x + 1 (x+2)2=y+3(x + 2)^2 = y + 3 (x+2)2=1(y+3)(x + 2)^2 = 1(y + 3)

So 4p=14p = 1, thus p=14p = \frac{1}{4}

Key Properties

  • Axis of symmetry passes through vertex
  • Focus lies on axis of symmetry
  • Distance from any point on parabola to focus = distance to directrix

📚 Practice Problems

1Problem 1easy

❓ Question:

Identify the vertex and axis of symmetry: y = (x - 3)² + 2

💡 Show Solution

Step 1: Recognize vertex form: y = (x - h)² + k where (h, k) is the vertex

Step 2: Identify values: h = 3, k = 2

Step 3: Vertex: (3, 2)

Step 4: Axis of symmetry: For vertical parabola: x = h x = 3

Answer: Vertex (3, 2), axis of symmetry x = 3

2Problem 2easy

❓ Question:

Find the vertex and focus of (x−1)2=8(y+2)(x - 1)^2 = 8(y + 2)

💡 Show Solution

This is in the form (x−h)2=4p(y−k)(x - h)^2 = 4p(y - k)

Comparing: (x−1)2=8(y−(−2))(x - 1)^2 = 8(y - (-2))

Vertex: (h,k)=(1,−2)(h, k) = (1, -2)

Find p: 4p=84p = 8, so p=2p = 2

Focus: (h,k+p)=(1,−2+2)=(1,0)(h, k + p) = (1, -2 + 2) = (1, 0)

Answer: Vertex (1,−2)(1, -2), Focus (1,0)(1, 0)

3Problem 3easy

❓ Question:

Find the focus and directrix of y² = 8x.

💡 Show Solution

Step 1: Recognize standard form: y² = 4px (horizontal parabola opening right)

Step 2: Identify 4p: 4p = 8 p = 2

Step 3: Find focus: For y² = 4px, focus is at (p, 0) Focus: (2, 0)

Step 4: Find directrix: For y² = 4px, directrix is x = -p Directrix: x = -2

Answer: Focus (2, 0), directrix x = -2

4Problem 4medium

❓ Question:

Write the equation of a parabola with vertex (0,0)(0, 0) and focus (3,0)(3, 0)

💡 Show Solution

The focus is to the right of the vertex, so this is a horizontal parabola opening right.

Use form: (y−k)2=4p(x−h)(y - k)^2 = 4p(x - h)

Vertex: (0,0)(0, 0) Focus: (3,0)(3, 0)

Since focus is at (h+p,k)=(3,0)(h + p, k) = (3, 0): p=3p = 3

4p=124p = 12

Equation: (y−0)2=12(x−0)(y - 0)^2 = 12(x - 0)

Answer: y2=12xy^2 = 12x

5Problem 5medium

❓ Question:

Convert to vertex form: y = x² - 6x + 5

💡 Show Solution

Step 1: Complete the square: y = x² - 6x + 5

Step 2: Take half of -6 and square it: (-6/2)² = (-3)² = 9

Step 3: Add and subtract 9: y = (x² - 6x + 9) - 9 + 5 y = (x - 3)² - 4

Step 4: Identify vertex: Vertex form: y = (x - 3)² - 4 Vertex: (3, -4)

Answer: y = (x - 3)² - 4

6Problem 6medium

❓ Question:

Write the equation of a parabola with vertex (2, 1) and focus (2, 3).

💡 Show Solution

Step 1: Determine orientation: Vertex (2, 1), Focus (2, 3) Same x-coordinate → vertical parabola

Step 2: Find p (distance from vertex to focus): p = 3 - 1 = 2

Step 3: Use vertex form: (x - h)² = 4p(y - k) where (h, k) is the vertex

Step 4: Substitute h = 2, k = 1, p = 2: (x - 2)² = 4(2)(y - 1) (x - 2)² = 8(y - 1)

Step 5: Find directrix: y = k - p = 1 - 2 = -1

Answer: (x - 2)² = 8(y - 1)

7Problem 7hard

❓ Question:

Find the vertex, focus, and directrix of y2−6y−4x+1=0y^2 - 6y - 4x + 1 = 0

💡 Show Solution

Step 1: Complete the square for yy y2−6y=4x−1y^2 - 6y = 4x - 1 y2−6y+9=4x−1+9y^2 - 6y + 9 = 4x - 1 + 9 (y−3)2=4x+8(y - 3)^2 = 4x + 8 (y−3)2=4(x+2)(y - 3)^2 = 4(x + 2)

Step 2: Identify vertex Vertex: (−2,3)(-2, 3)

Step 3: Find pp 4p=44p = 4, so p=1p = 1

Step 4: Find focus (horizontal parabola) Focus: (h+p,k)=(−2+1,3)=(−1,3)(h + p, k) = (-2 + 1, 3) = (-1, 3)

Step 5: Find directrix Directrix: x=h−p=−2−1=−3x = h - p = -2 - 1 = -3

Answer: Vertex (−2,3)(-2, 3), Focus (−1,3)(-1, 3), Directrix x=−3x = -3

8Problem 8hard

❓ Question:

A satellite dish is shaped like a paraboloid. If the dish is 8 feet across and 2 feet deep, where should the receiver be placed?

💡 Show Solution

Step 1: Set up coordinate system: Place vertex at origin (0, 0) Parabola opens upward: x² = 4py

Step 2: Identify a point on the parabola: At the edge: x = 4 (half of 8), y = 2 Point: (4, 2)

Step 3: Substitute to find p: 4² = 4p(2) 16 = 8p p = 2

Step 4: Find the focus: Focus is at (0, p) = (0, 2)

Step 5: Interpret: The receiver should be at the focus, 2 feet above the vertex

Answer: 2 feet from the bottom, at the center

🎴

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