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Ellipse Tangent


EllipseNormal

The normal to an ellipse at a point P intersects the ellipse at another point Q. The angle corresponding to Q can be found by solving the equation

 (P-Q)·(dP)/(dt)=0
(1)

for t^', where P(t)=(acost,bsint) and Q(t)=(acost^',bsint^'). This gives solutions

 t^'=+/-cos^(-1)[+/-(N(t))/(a^4sin^2t+b^4cos^2t)],
(2)

where

 N(t)=1/2b^2cost[a^2+b^2+(b^2-a)^2cos(2t)]+a^2(a-b)(a+b)costsin^2t,
(3)

of which (+,-) gives the valid solution. Plugging this in to obtain Q then gives

d(t)=|P-Q|
(4)
=(sqrt(2)ab[a^2+b^2+(b^2-a^2)cos(2t)]^(3/2))/(a^4+b^4+(b^4-a^4)cos(2t))
(5)
=(2ab(b^2cos^2t+a^2sin^2t)^(3/2))/(b^4cos^2t+a^4sin^2t).
(6)

To find the maximum distance, take the derivative and set equal to zero,

 d^'(t)=(2ab(a-b)(a+b)costsintsqrt(b^2cos^2t+a^2sin^2t))/((b^4cos^2t+a^4sin^2t)^2)×(a^4sin^2t+b^4cos^2t-2a^2b^2)=0,
(7)

which simplifies to

 a^4sin^2t+b^4cos^2t-2a^2b^2=0.
(8)

Substituting for sin^2t and solving gives

cos^2t=(a^4-2a^2b^2)/(a^4-b^4)
(9)
sin^2t=(2a^2b^2-b^4)/(a^4-b^4).
(10)

Plugging these into d(t) then gives

 d_(min)=(3sqrt(3)a^2b^2)/((a^2+b^2)^(3/2)).
(11)

This problem was given as a Sangaku problem on a tablet from Miyagi Prefecture in 1912 (Rothman 1998). There is probably a clever solution to this problem which does not require calculus, but it is unknown if calculus was used in the solution by the original authors (Rothman 1998).


See also

Ellipse

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References

Rothman, T. "Japanese Temple Geometry." Sci. Amer. 278, 85-91, May 1998.

Referenced on Wolfram|Alpha

Ellipse Tangent

Cite this as:

Weisstein, Eric W. "Ellipse Tangent." From MathWorld--A Wolfram Web Resource. https://mathworld.wolfram.com/EllipseTangent.html

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