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)#