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Bezier equation and 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.

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.

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)

✓ Answer checked by Claude Opus 5.5 (high effort)

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