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### Minimum Distance Between a Line and an Ellipse

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Date: 05/30/2000 at 09:45:27
From: Alexander Hajenius
Subject: Minimum distance between line & ellipse

I have to find the extremum from a point on the line x + y = 4 to a
point on the ellipse x^2 + 4 y^2 = 4. In other words, what's the
minimum distance from the line to the ellipse? I know the formula for
calculating the distance between the two points but I cannot find the
function that should be minimized.

Also, is there a general way to solve such problems in a 3-dimensional
space?

Alexander Hajenius
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Date: 05/30/2000 at 16:51:10
From: Doctor Rob
Subject: Re: Minimum distance between line & ellipse

Thanks for writing to Ask Dr. Math, Alexander.

First of all, you have a problem with the word "extremum." It means
either maximum or minimum, so you are seeking both.

The points where the extrema are located are points where the tangent
to the ellipse is parallel to the line. The slope of the line is -1,
so you want to find the points where the tangent has slope -1. Do this
by using implicit differentiation on the equation of the ellipse:

2*x + 8*y*(dy/dx) = 0
dy/dx = -x/(4*y)
-x/(4*y) = -1
x = 4*y

The extrema both lie on this line and the ellipse, so finding their
intersection will give you the extrema. One will be a maximum and one
a minimum, and computing the actual maximum and minimum distance isn't
very hard. The distance from a point (x0,y0) to the line
A*x + B*y + C = 0 is given by:

d = |A*x0+B*y0+C|/sqrt(A^2+B^2)

In three dimensions, the situation is similar. The extrema of the
distance from a plane to an ellipsoid are the points of tangency where
the tangent plane is parallel to the given plane. These can be found
by partial differentiation of the equations of the surface and of the
plane. The partials with respect to each variable should be
proportional one to the other. That will give you two equations of
planes, whose intersection is a line on which the two extrema lie, as
well as lying in the surface. Solving these three equations together
will give you the coordinates of the two extrema, and the distance
from a point (x0,y0,z0) to the line A*x + B*y + C*z + D = 0 is given
by:

d = |A*x0+B*y0+C*z0+D|/sqrt(A^2+B^2+C^2)

- Doctor Rob, The Math Forum
http://mathforum.org/dr.math/
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Associated Topics:
College Calculus

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