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Parabolas (Conic Form)

Parabolas with horizontal and vertical orientations

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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:

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)

2Problem 2medium

❓ 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

3Problem 3hard

❓ 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

Explain using:

⚠️ Common Mistakes: Parabolas (Conic Form)

Avoid these 3 frequent errors

🌍 Real-World Applications: Parabolas (Conic Form)

See how this math is used in the real world

📝 Worked Example: Solving a Quadratic by Factoring

Problem:

Solve x2−5x+6=0x^2 - 5x + 6 = 0.

2Factor the quadratic
3Set each factor equal to zero
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❓ Frequently Asked Questions

What is Parabolas (Conic Form)?▾
Parabolas with horizontal and vertical orientations
How can I study Parabolas (Conic Form) effectively?▾
Start by reading the study notes and working through the examples on this page. Then use the flashcards to test your recall. Practice with the 3 problems provided, checking solutions as you go. Regular review and active practice are key to retention.
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What course covers Parabolas (Conic Form)?▾
Parabolas (Conic Form) is part of the Algebra 2 course on Study Mondo, specifically in the Conic Sections section. You can explore the full course for more related topics and practice resources.
Are there practice problems for Parabolas (Conic Form)?▾
Yes, this page includes 3 practice problems with detailed solutions. Each problem includes a step-by-step explanation to help you understand the approach.