 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Minimize Angle (Posted on 2013-11-30) 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 No Rating 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) Therefore,    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. Construction: 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). QED Comments: ( You must be logged in to post comments.)
 Subject Author Date Direct calculus approach Jer 2013-11-30 14:51:24 Please log in:

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