The normal to an ellipse at a point 
intersects the ellipse at
another point 
.
The angle corresponding to 
can be found by solving the equation
 |
(1)
|
for 
,
where 
and 
.
This gives solutions
![t^'=+/-cos^(-1)[+/-(N(t))/(a^4sin^2t+b^4cos^2t)],](/images/equations/EllipseTangent/NumberedEquation2.svg) |
(2)
|
where
![N(t)=1/2b^2cost[a^2+b^2+(b^2-a)^2cos(2t)]+a^2(a-b)(a+b)costsin^2t,](/images/equations/EllipseTangent/NumberedEquation3.svg) |
(3)
|
of which 
gives the valid solution. Plugging this in to obtain 
then gives
 |  |  |
(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))](/images/equations/EllipseTangent/Inline14.svg) |
(5)
|
 |  |  |
(6)
|
To find the maximum distance, take the derivative and set equal to zero,
 |
(7)
|
which simplifies to
 |
(8)
|
Substituting for 
and solving gives
 |  |  |
(9)
|
 |  |  |
(10)
|
Plugging these into 
then gives
 |
(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).
