The ideas explained below are based in the hypothesis that the involved functions are twice continuous.
The problem can be stated as: given a set of points (x,y) defined through the relationship f(x,y) = 4x^2+y^2-4=0, and a fixed point q=(x_q,y_q), determine a point p_o=(x_o, y_o) in the set defined by f(x,y)=0 such that the distance d(x,y,x) = norm(p_o-q) has maximum value.
If such point exists, then the normal to f(x,y)=0 in p_o must be linearly dependent with the normal to d(x,y,z) in p_o. In other words, the two normals must be co-lineal. Formally this condition is equivalent to
grad f(x_o,y_o) = lambda grad d(x_o,y_o)
where grad = (partial/(partial x),partial/(partial x))
or
{(f_x = lambda d_x),(f_y = lambda d_y):}
or
{(8x=lambda (x-x_q)/(d(x,y))), (2y=lambda (y-y_q)/ (d(x,y))):}
or
{(8x=lambda_d (x-x_q)), (2y=lambda_d (y-y_q)):}
here lambda_d = lambda/(d(x,y))
Putting all together we have the conditions:
{(4x_o^2+y_o^2-4=0),(8x_o=lambda_d (x_o-x_q)), (2y_o=lambda_d (y_o-y_q)):}
Now making
p_q=(x_q,y_q) = (1,0) as proposed
and solving for x_o,y_o,lambda_d we have
{
(x = -1., y = 0., lambda_d = 4.),
(x = -0.333333, y = -1.88562, lambda_d = 2.),
(x = -0.333333, y = 1.88562, lambda_d= 2.)
:}
The computed distances are respectively
(2., 2.3094, 2.3094)
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