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`4sqrt(11)``2sqrt(11)``4sqrt(22)``2sqrt(22)`

Answer :

DSolution :

Centre `-=(1,2)` <br> Radius of auxiliary circle `=a =sqrt((2-1)^(2)+(5-2)^(2))=sqrt(10)` <br> `2ae=sqrt(8^(2)+8^(2))=8sqrt2 or e=(4)/(sqrt5)` <br> `b^(2)=a^(2)e^(2)-a^(2)=32-10=22` <br> `"or "2b=2sqrt(22)`