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If radii of director circles of `x^2/a^2+y^2/b^2=1 and x^2/a^2-y^2/b^2=1` are `2r and r` respectively, let `e_E and e_H` are the eccentricities of ellipse and hyperbola respectively, then

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The correct Answer is:
6
Equation of director cicles of ellipse and hyperbola are, respectively,
`x^(2)+y^(2)=a^(2)+b^(2)`
`and x^()+y^(2)=a^(2)-b^(2)`
According to the question, we have
`a^(2)+b^(2)=4r^(2)" (1)"`
`a^(2)-b^(2)=r^(2)" (2)"`
Solving Eqs. (1) and (2), we get
`a^(2)=(5r^(2))/(2),b^(2)=(3r^(2))/(2)`
`e_(1)^(2)=1-(b^(2))/(a^(2))`
`=1-(3)/(5)=(2)/(5)`
`e_(2)^(2)=1+(b^(2))/(a^(2))`
`=1+(3)/(5)=(8)/(5)`
So, `4e_(2)^(2)-e_(1)^(2)=4xx(8)/(5)-(2)/(5)=6`
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