Minimum Distance Between a Line and an EllipseDate: 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? Please help, Alexander Hajenius 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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