Lagrange Multipliers help!?

It's easier to minimize the square of the distance from (1, 0), which is the same as minimizing the distance.

The square of the distance is given by the function f(x,y) = (x-1)^2 + y^2.

The ellipse has constraint equation g(x,y) = x^2/4 + y^2 = 1.

Since grad f = lambda*grad g, we have the equations

1) 2(x-1) = lambda*x/2

2) 2y = lambda*2y

3) x^2/4 + y^2 = 1.

Because 2) is equivalent to 2y(1-lambda) = 0, y = 0 or lambda = 1.

Case 1: lambda = 1

Then from 1), we have 2(x-1) = x/2; 2x - x/2 = 2; x = 4/3.

Then from 3), (4/3)^2 / 4 + y^2 = 1; 4/9 + y^2 = 1; y = +-(sqrt 5)/3.

So we get critical points (4/3, (sqrt 5)/3) and (4/3, -(sqrt 5)/3).

Case 2: y = 0

Then from 3), x^2/4 = 1 and so x = +-2.

Note that for each of these x values, some value of lambda will satisfy 1), since neither x value makes x/2 equal to zero.

So we get additional critical points (2, 0) and (-2, 0).

Now we compare values of f(x, y) = (x-1)^2 + y^2 at all the critical points.

f(2, 0) = 1

f(-2, 0) = 9

f(4/3, (sqrt 5)/3) = 2/3.

f(4/3, -(sqrt 5)/3) = 2/3.

So the distance from (1, 0) is minimized at the points (4/3, +-(sqrt 5)/3) on the ellipse, and this minimum distance equals sqrt[(4/3, +-(sqrt 5)/3)] = sqrt(2/3) = (sqrt 6)/3.

Lord bless you today!