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Determining Tangent to an Ellipse, Minimizing Area

Date: 02/20/98 at 09:14:01
From: Adam Huesser
Subject: calculus 

Find the equation of the tangent to the ellipse

   b^2*x^2+a^2*y^2 = a^2*b^2

in the first quadrant that forms with the coordinate axes the triangle 
of the smallest possible area (a & b are positive constants).

Date: 02/20/98 at 17:00:19
From: Doctor Rob
Subject: Re: calculus 

Let the equation of the line be x/A + y/B = 1. A is the x-intercept 
and B is the y-intercept of the line. Since we are working in the 
first quadrant, A > 0 and B > 0. Being tangent to the ellipse at 
(x0,y0) means that these coordinates satisfy the equation of the 
ellipse, the equation of the line, and the slope of the line equals 
(dy/dx) evaluated at (x0,y0), which is the slope of the curve. 
Furthermore, x0 > 0 and y0 > 0. Then

   b^2*x0^2 + a^2*y0^2 = a^2*b^2,
   x0/A + y0/B = 1,
   -B/A = dy/dx = -b^2*x0/(a^2*y0).

Subject to these constraints, you want to minimize the area S = A*B/2.

Solve the second equation for B in terms of A, x0, and y0. Substitute
that in the third equation and the expression for S. Use the new third 
equation to solve for A as a function of x0 and y0. Substitute that in 
the new second equation and the expression for S. Now use the first 
equation to express y0 in terms of x0, and substitute that in all the 
other equations. During this process, you may need to use the facts 
that A, B, x0, and y0 are all positive.

Now you have S as a function of x0 alone. Compute dS/dx0, set it equal 
to zero, and solve for x0. Among those roots will be the one which 
minimizes S. Use that value to figure out A and B, and substitute them 
into the equation of the line.

-Doctor Rob, The Math Forum
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Associated Topics:
High School Calculus
High School Equations, Graphs, Translations

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