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A. `e_(1)^(2)+e_(2)^(2)=(43)/(40)`
B. `e_(1)e_(2)=(sqrt(7))/(2sqrt(10))`
C. `|e_(1)^(2)-e_(2)^(2)|=(5)/(8)`
D. `e_(1)e_(2)=(sqrt(3))/(7)`

Explanation

Let ellipse be `E_(1):(x^(2))/(a^(2))+(y^(2))/(b^(2))=1 and E_(2)=(x^(2))/(A^(2))+(y^(2))/(B^(2))=1`
Sincex+y-3 is a tangent,
`a^(2)+b^(2)=A^(2)+b^(2)=9` (using condition `c^(2)=a^(2)m^(2)+b^(2)` etc).
Point P lies on `x^(2)+(y-1)^(2)=2`
Euation of normal to circle having slop 1 is
`y-1=1x(x-0) or y-y+1=0`
Solving this normal with tangent line we get point P(1,2)
Now, `PA=PR=(2sqrt(2))/(3)`
So, points on line `x+y-3=0` at distance `(2sqrt(2))/(3)` from point
`P "are"(1+-(1)/(sqrt(2))(2sqrt(2))/(3),2+-(1)/(sqrt2)(2sqrt(2))/(3))`
or `Q((5)/(3),(4)/(3)) and Q ((1)/(3),(8)/(3))`
Now, `Q((5)/(3),(4)/(3))` lies on `E_(1)`
So, `(25)/(9a^(2))+(16)/(9(9-a^(2)))=1`
`rArr2525-25a^(2)+16a^(2)=9a^(2)(9-a^(2))`
`rArra^(2)-10a^(2)+25=0`
`rArr a^(2)=5 so b^(2)=4`
`:. e_(1)^(2)=1(b^(2))/(a^(2))=1-(4)/(5)=(1)/(5)`
Now, `((1)/(3),(8)/(3))` lies on `E_(2)`
So, `(1)/(A^(2))+(64)/((9-A^(2))=9`
`rArr9-A^(2)+64A^(2)=9A^(2)(9-A^(2))`
`rArr-2A^(2)+1=0`
`rArr A^(2)=1 " " so, B^(2)=8`
`:. =1- =(1)/(8)=(7)/(8)`