Solution: Quadratic Constraints in a Plane

Return to current section: Parametric Equations.


Given the plane x(α,β) = α+ β+ c, sketch curves for the quadratic constraints listed below


          1. 
αβ1  {Note: cos2(θ) + sin2(θ) = 1}

          2. β α2 

          3. α− β1  {Note: cosh2(λ) - sinh2(λ) = 1}

   
Simple symmetrical solutions:


          
1. 
x(θcos(θ) + sin(θ) + c  {Parametric ellipse, x(0) a + c}

          2. x(αaα + bα2 c  {Parametric parabola, x(0) c}

          3. x(λcosh(λ) + sinh(λ) + c  {Parametric hyperbola, x(0) a + c}


In the interactive diagram below, each of these three curves has its own slider for varying its parameter. The corresponding position vector traces out its curve as the slider is adjusted. To reset (clearing the traces to focus on another slider), click the icon in the top right corner of the diagram (sometimes the icon is not visible, but clicking the mouse in that corner should still work).




We've chosen 
c = (2,1,0) drawn in greena = (2,0,0) drawn in blue, and b = (1,1,0) drawn in red. In general, vectors a and b determine the plane to which c reaches from the coordinate origin. Notice that since a and b are not perpendicular, neither vector c nor vector a+c points to the vertex of any of the curves. 


For the ellipse,
dx/dθ = b cos(θ) - a sin(θ) is a tangent vector to the curve and at θ=0 vector b is parallel to this tangent. In the demo above, move the black slider to θ=0 and check this statement graphically.


For the hyperbola, dx/dλ = a sinh(λ) + b cosh(λ) is a tangent vector to the curve and at λ=0 vector b is again parallel to this tangent. In the demo above move the orange slider to λ=0 and check this statement graphically.


Considering the last two paragraphs, notice further that θ=0 and 
λ=0 are where the ellipse and hyperbola intersect each other, are tangent to each other, and therefore share b as a common tangent vector.


For the parabola, dx/d
α = a + 2αb is a tangent vector to the curve and at α=0 this time vector a is parallel to this tangent. It is interesting to observe in the demo (although not so easy to prove algebraically) that vector b is parallel to the axis of symmetry, even though it doesn't pass through the vertex. 




Return to current section
Parametric Equations.



© David Hestenes 2005, 2014