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| Minimize Angle (Posted on 2013-11-30) |
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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.
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Submitted by Bractals
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Solution:
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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
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