If `P(alpha, beta)`, the point of intersection of the ellipse `x^2/a^2+y^2/a^2(1-e^2)=1` and hyperbola `x^2/a^2-y^2/(a^2(E^2-1)=1/4`is equidistant from the foci of the curvesall lying in the right of y-axis thenA. `2 alpha =a (2e +E)`B. `a- ealpha = E alpha -alpha//2`C. `E =(sqrt(e^(2)+24)-3e)/(2)`D. `E=(sqrt(e^(2)+12)-3e)/(2)`

If `P(alpha, beta)`, the point of intersection of the ellipse `x^2/a^2

1.	If `P(alpha, beta)`, the point of intersection of the ellipse `x^2/a^2+y^2/a^2(1-e^2)=1` and hyperbola `x^2/a^2-y^2/(a^2(E^2-1)=1/4`is equidistant from the foci of the curvesall lying in the right of y-axis thenA. `2 alpha =a (2e +E)`B. `a- ealpha = E alpha -alpha//2`C. `E =(sqrt(e^(2)+24)-3e)/(2)`D. `E=(sqrt(e^(2)+12)-3e)/(2)`
Answer» Correct Answer - B::C Focus of ellipse lying to the right of the y-axis is `S_(1) (ae,0)` Focus of hyperbola lying to the right of the y-axis is `S_(2)(aE//2,0)` Now `P(alpha, beta)` is equidistance from `S_(1)` and `S_(2)` `:. S_(1)P = S_(2)P rArr a - e alpha = E alpha -((a)/(2))`. (1) Also P lies on perpendicular bisector of `S_(1)S_(2)` `:. alpha = (ae+(a)/(2)E)/(2)` From (1) and (2), `E^(2) + 3eE + (2e^(2)-6) =0` `rArr E = (sqrt(e^(2)+24)-3e)/(2)`

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