Focii of the ellipse are at # f1 (0,0) and f2 (1,1)#
Distance between them #f1f2= sqrt ((x_1-x_2)^2+(y_1-y_2)^2#
or #f1f2= sqrt((0-1)^2+(0-1)^2) =sqrt2#
Distance of the focii from centre is #c=sqrt2/2#
Semi major axis length #a=4/2=2# The relation of
#c, a, b# is # c^2 = a^2 - b^2:. (sqrt2/2)^2=2^2-b^2 # or
#b^2=4-1/2 or b=sqrt 3.5 #. The center is #(h=1/2,k=1/2)#
The ellipse is tilted at an angle of #theta=tan^-1 (1/1)=45^0#
Hence the equation of ellipse is.
#{(x-h)cos theta+(y-k)sin theta}^2/a^2+{(y-k)cos theta-(x-h)sin theta}^2/b^2 =1# or
#{(x-1/2)cos 45+(y-1/2)sin 45}^2/4+{(y-1/2)cos 45-(x-1/2)sin 45}^2/3.5 =1# or
#{(x-1/2)/sqrt2+(y-1/2)/sqrt2}^2/4+{(y-1/2)/sqrt2-(x-1/2)/sqrt2}^2/3.5 =1#
#{(2x-1)/(2sqrt2)+(2y-1)/(2sqrt2)}^2/4+{(2y-1)/(2sqrt2)-(2x-1)/(2sqrt2)}^2/3.5 =1# or
#{(2x-1+2y-1)/(2sqrt2)}^2/4+{(2y-2x)/(2sqrt2)}^2/3.5 =1# or
#(x+y-1)^2/8+(y-x)^2/7 =1# [Ans]
