Ellipse Tangent
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
| [画像: (P-Q)·(dP)/(dt)=0 ] |
(1)
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for t^', where P(t)=(acost,bsint) and Q(t)=(acost^',bsint^'). This gives solutions
![画像: t^'=+/-cos^(-1)[+/-(N(t))/(a^4sin^2t+b^4cos^2t)],](/index.cgi/contrast/https://mathworld.wolfram.com/images/equations/EllipseTangent/NumberedEquation2.svg){kind=link}
where
| N(t)=1/2b^2cost[a^2+b^2+(b^2-a)^2cos(2t)]+a^2(a-b)(a+b)costsin^2t, |
(3)
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of which (+,-) gives the valid solution. Plugging this in to obtain Q then gives
![画像:(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))](/index.cgi/contrast/https://mathworld.wolfram.com/images/equations/EllipseTangent/Inline14.svg){kind=link}
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)
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Substituting for sin^2t and solving gives
Plugging these into d(t) then gives
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
EllipseExplore with Wolfram|Alpha
More things to try:
References
Rothman, T. "Japanese Temple Geometry." Sci. Amer. 278, 85-91, May 1998.Referenced on Wolfram|Alpha
Ellipse TangentCite this as:
Weisstein, Eric W. "Ellipse Tangent." From MathWorld--A Wolfram Resource. https://mathworld.wolfram.com/EllipseTangent.html