---
title: "Bezier equation and point at u=0.8"
subject: "Computer Graphics"
chapter: 5
key: "semester_5/computer_graphics:bezier-equation-point-u-08"
source: https://www.bctengineer.com/q/semester_5/computer_graphics/bezier-equation-point-u-08
site: BCT Engineering
---

# Bezier equation and point at u=0.8

Computer Graphics · Chapter 5: Curve Modeling

- Asked in: 2071 Chaitra, 10 marks
- Video: [Bezier Curve Examples in computer graphics | Bezier curve Numerical | Lec-38](https://www.youtube.com/watch?v=tXiQ5ESEAhA) — Er Sahil ka Gyan
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/computer_graphics/bezier-equation-point-u-08
- Whole chapter: https://www.bctengineer.com/md/semester_5/computer_graphics/chapter-5

## Question

Write down properties of Bezier curve. Find equation of Bezier curve whose control points are P0(2,6), P1(6,8) and P2(9,12). Also find co-ordinate of point at u = 0.8.

## Short answer

For P0(2,6), P1(6,8), P2(9,12), the quadratic Bezier is x(u)=-u^2+8u+2, y(u)=2u^2+4u+6; at u=0.8 it gives (7.76, 10.48). Properties include passing through first/last control points, degree one less than point count, convex hull containment, affine invariance, global control, and symmetry.

## Answer

#### Properties
1. Passes through the first and last control points, P(0) = P0, P(1) = Pn.
2. Degree = number of control points − 1.
3. End tangents lie along the first and last control polygon edges.
4. The curve lies inside the convex hull of the control points.
5. Blending functions are positive and add to 1.
6. Affine invariance.
7. Global control (no local control).
8. Symmetric with respect to point order.

#### Equation (three points, so n = 2, quadratic)
P(u) = (1−u)² P0 + 2u(1−u) P1 + u² P2

Expanding: P(u) = (P0 − 2P1 + P2) u² + (−2P0 + 2P1) u + P0

- P0 − 2P1 + P2 = (2 − 12 + 9, 6 − 16 + 12) = (−1, 2)
- −2P0 + 2P1 = (−4 + 12, −12 + 16) = (8, 4)
- P0 = (2, 6)

**x(u) = −u² + 8u + 2, y(u) = 2u² + 4u + 6**, 0 ≤ u ≤ 1

Check: u = 0 gives (2,6); u = 1 gives (9,12), as required.

#### Point at u = 0.8
Blending values: B0 = (0.2)² = 0.04, B1 = 2(0.8)(0.2) = 0.32, B2 = (0.8)² = 0.64 (sum = 1)

x = 0.04·2 + 0.32·6 + 0.64·9 = 0.08 + 1.92 + 5.76 = 7.76

y = 0.04·6 + 0.32·8 + 0.64·12 = 0.24 + 2.56 + 7.68 = 10.48

(Using the equation: x = −0.64 + 6.4 + 2 = 7.76, y = 1.28 + 3.2 + 6 = 10.48)

**Answer: P(0.8) = (7.76, 10.48)**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Three control points P0, P1, P2 give a Bezier curve of degree:
   A. 3
   B. 2 (quadratic)
   C. 1
   D. 4
   Answer: B. Degree = number of control points - 1 = 2. (Key point: Degree)

2. Which is the quadratic Bezier equation?
   A. (1-u)^2 P0 + 2u(1-u) P1 + u^2 P2
   B. (1-u)^2 P0 + u(1-u) P1 + u^2 P2
   C. (1-u)^3 P0 + 3u(1-u)^2 P1 + u^3 P2
   D. (1-u) P0 + 2u P1 + (1-u)^2 P2
   Answer: A. Coefficients C(2,i) = 1, 2, 1. (Key point: Quadratic formula)

3. For P0(2,6), P1(6,8), P2(9,12), x(u) equals:
   A. u^2 + 8u + 2
   B. -u^2 + 8u + 2
   C. -u^2 + 4u + 2
   D. 2u^2 + 4u + 6
   Answer: B. x: P0-2P1+P2 = 2-12+9 = -1, -2P0+2P1 = 8, P0 = 2. (Key point: Expanded x(u))

4. What is P(0.8)?
   A. (7.76, 8.48)
   B. (6.12, 10.48)
   C. (9, 12)
   D. (7.76, 10.48)
   Answer: D. Blending 0.04, 0.32, 0.64 give x=7.76 and y=10.48. (Key point: Point at u=0.8)

## Related in Computer Graphics

- [Draw Bezier curve through 4 control points](https://www.bctengineer.com/question.md?key=semester_5%2Fcomputer_graphics%3Abezier-four-points-p0-0-0) — Computer Graphics, chapter 5
- [Bezier properties; point at u=0.2](https://www.bctengineer.com/question.md?key=semester_5%2Fcomputer_graphics%3Abezier-properties-point-u-0-2) — Computer Graphics, chapter 5, asked 2×
- [Derive cubic Bezier equation and properties](https://www.bctengineer.com/question.md?key=semester_5%2Fcomputer_graphics%3Abezier-cubic-equation-properties) — Computer Graphics, chapter 5
- [Bezier curve through given points](https://www.bctengineer.com/question.md?key=semester_5%2Fcomputer_graphics%3Abezier-two-properties-curve-through-points) — Computer Graphics, chapter 5
- [Bezier curve and its properties](https://www.bctengineer.com/question.md?key=semester_5%2Fcomputer_graphics%3Abezier-curve-properties) — Computer Graphics, chapter 5, asked 6×
- [Liang-Barsky: P1(-5,3), P2(15,9)](https://www.bctengineer.com/question.md?key=semester_5%2Fcomputer_graphics%3Aliang-barsky-0-0-10-10) — Computer Graphics, chapter 3

## Related in other subjects

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- [Bezier surface equation and properties](https://www.bctengineer.com/question.md?key=semester_4%2Fcomputer_aided_design%3Abezier-surface-properties) — Computer Aided Design, chapter 4
- [Cubic Bezier point and tangent calculation](https://www.bctengineer.com/question.md?key=semester_4%2Fcomputer_aided_design%3Abezier-curve-numerical) — Computer Aided Design, chapter 3