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Minimize Angle (Posted on 2013-11-30) Difficulty: 3 of 5

Let Γ be the ellipse ( b2x2 + a2y2 = a2b2 with a > b ). Let O
denote the origin.

Construct a line tangent to Γ ( at point P in the 1st quadrant )
and intersecting the y-axis at point Q such ∠QPO is minimized.

  Submitted by Bractals    
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Solution: (Hide)

Let the tangent line PQ intersect the x-axis at point R.

   tan(∠PRO) = - slope(PQ) = (b2/a2)(x/y) = b2/a2tan(∠ROP)


   tan(∠QPO) = tan(∠ROP + ∠PRO)
         = [tan(∠ROP) + tan(∠PRO)]/[1 - tan(∠ROP)tan(∠PRO)]
         = [a2tan2(∠ROP) + b2]/ (a2 - b2)tan(∠ROP)

Therefore, the miminmum occurs when

   [(a2 - b2)tan(∠ROP)][a2tan2(∠ROP) + b2]' -
   [a2tan2(∠ROP) + b2][(a2 - b2)tan(∠ROP)]' = 0,
   where the prime means differentiation with respect to ∠ROP.

Which is tan(∠ROP) = b/a and the slope(PQ) = -b/a.


P is the intersection of the line (0,0)-(a,b) and Γ.

The tangent line is through P and parallel to the line (0,b)-(a,0).


Comments: ( You must be logged in to post comments.)
  Subject Author Date
SolutionDirect calculus approachJer2013-11-30 14:51:24
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