1.

Let E_(1) and E_(2) two ellipse whose centres are at the orgin. Then major axes of E_(1) and E_(2) lie along the x-axis and the y-axis, respectively. LetS be the circle x^(2)+(y-1)^(2)=2 the straight line x + y = 3 touches the curves S, E_(1) and E_(2) at P, Q and R, respectively. Suppose that PQ = PR = (2sqrt2)/(3), if e_(1) and e_(2) are the eccentricities of E_(1) and E_(2), respectively, then the correct expression(s) is (are)

Answer»

`e_(1)^(2)+e_(2)^(2)=(43)/(40)`
`e_(1)e_(2)=(SQRT7)/(2sqrt10)`
`|e_(1) ^(2)-e_(2)^(2)|=(5)/(8)`
`e_(1)e_(2)=(sqrt3)/(4)`

Solution :Let the equations of ellipses `E_(1) and E_(2)` be `(x^(2))/(a^(2))+(y^(2))/(B^(2))=1 and (x^(2))/(A^(2))+(y^(2))/(B^(2))=1` respectively. Then,
`e_(1)^(2)=1(b^(2))/(a^(2))and e_(2)^(2)=1(A^(2))/(B^(2))`
The slope of the x+y = 3 is -1. It TOUCHES ellipses `E_(1) and E_(2)`. Therefore, `a^(2)+b^(2)=3 and A^(2)+B^(2)=3`. The COORDINATES of Q and R are `((A^(2))/(3),b^(2)/(3))and((A^(2))/(3),(B^(2))/(3))` respectively.
The equation of the line passing through the centre (0, 1) of the circle and perpendicular to `x+y=3 " is " y-1=1(x-0) or x-y+1=0`. This line intersects `x+y=3` at (1, 2). Thus, the line `x+y=3` touches the circle S at (1,2). The equation of the line`x+y=3` in distance form is
`(x-1)/(cos3pi/4)=(y-2)/(sin3pi//4)or(x-1)/(-1//sqrt2)=(y-2)/(1//sqrt2)`
The coordinates of Q and R are given by
`(x-1)/(-1//sqrt2)=(y-2)/(1//sqrt2)=(2sqrt2)/(3)and, (x-1)/(-1//sqrt2)=(y-2)/(1//sqrt2)=(2sqrt2)/(3)` respectively.
The coordinates of Q and R are `(5//3,4//3)and(1//3, 8//3)` respectively.
but, the coordinates of Q and R are `(a^(2)//3, b^(2)//3) and (A^(2)//3,B^(2)//3)`respectively.
`therefore (a^(2))/(3)=(5)/(3),(b^(2))/(3)=(4)/(3),(A^(2))/(3)=(1)/(3) and (B^(2))/(3)=(8)/(3)`
`a^(2)=5,b^(2)=4, A^(2)=1 and B^(2)=8`
`therefore e_(1)^(2)=1-(4)/(5)and e_(1)^(2)=1-(1)/(8)=(7)/(8)`
`rArr e_(1)^(2)+e_(2)^(2)=(43)/(40)and e_(1)e_(2)=(sqrt7)/(2sqrt10)`


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