From my calculus class

[BMAD]

Here is a calculus problem from a class i teach. The problem itself illustrates the benefits of recognizing the fluidity and openness one can take in mathematics as the direct approach is ugly and messy but there is a simpler and elegant indirect way of solving this one too. Enjoy.

Find the equation of the line 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 smallest possible area (a & b are positive constants)

[bonanova]

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The answer can be found fairly quickly by performing an affine transformation of coordinates.

Divide the equation by a²b².

Define u=x/a and v=y/b so that u²+v²=1.

1. The tangent line at v=u forms the minimal triangle.
2. Substitute back to get the equation in terms of x and y.

Show the tangent at u=v gives the minimum area:

1. The u=v tangent is v = sqrt(2) - u with intercepts u_i = v_i = sqrt (2).
   A _(u=v) = 1/2 u_i v_i = 1. Show this is minimal. 
   .....
2. A tangent at arbitrary u in (0,1) gives A_u,v = 1/2uv; A_u = 1/(2u sqrt(1-u²)).
   A_u is infinite at 0 and 1; its extremum in (0,1) is thus a minimum.
   Let f = 1/2A. f_u,v = uv; f_u = u sqrt(1-u²). df_u,v/du = v - u²/v.  df/du = 0 iff u=v. 

Express the u=v tangent in x-y coordinates.

1. u=x/a and v=y/b
2. (y/b) = sqrt(2) - (x/a) => y = (b/a)(a sqrt(2) - x)

Affine transformations preserve relative areas; the area is thus minimized in both planes.

[bonanova]

The answer can be found fairly quickly by performing an affine transformation of coordinates. Divide the equation by a²b². Define u=x/a and v=y/b so that u² + v² = 1. The tangent line at v=u forms the minimal triangle. Substitute back to get the equation in terms of x and y.

 

This works, because affine transformations preserve (relative) areas, and the equation in u-v space is much easier to derive.