First, if you maximise d you also maximise d^2, using d^2 avoids that

sqrt so makes everything easier, so we want to maximise:

D2 = d^2 = (x-1)^2 + (y-0)^2 = (x-1)^2 + y^2

but from the equation for the ellipse:

y^2 = 4 - 4x^2

So:

D2 = (x-1)^2 - 4(x^2 -1) = - 3 x^2 - 2 x + 5

Now:

d(D2)/dx = -6 x -2

for a maximum of D2 d(D2)/dx=0, so x=-1/3, and so D2 = 16/3, and y=4sqrt(2)/3

(assuming I have done the algebra correctly).

RonL