InterviewSolution
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InterviewSolution
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Let E_(1) and E_(2)be two ellipse whsoe centers are at the origin. The major axes of E_(1) and E_(2) lie along the x-axis , and the y-axis, respectively. Let S be the circle x^(2)+(y-1)^(2)=2 . The straigth line x+y=3 touches the curves, S, E_(1) and E_(2) at P,Q and R, respectively . Suppose that PQ=PR=(2sqrt(2))/(3). If e_(1) and e_(2) are the eccentricities of E_(1) and E_(2) respectively, thent hecorrect expression (s) is (are) |
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Answer» `e_(1)^(2)+e_(2)^(2)=(43)/(40)` 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)` |
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