Determining Tangent to an Ellipse, Minimizing AreaDate: 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 Check out our web site http://mathforum.org/dr.math/ |
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