If a variable line has its intercepts on the coordinates axes as e,e′ where e2,e′2 are the eccentricities of a hyperbola and its conjugate hyperbola respectively, then the line always touches the circle centred at O whose radius r is equal to
A
2
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B
2.0
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C
2.00
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Solution
Since, e2 and e′2 are the eccentricities of a hyperbola and its conjugate hyperbola respectively. ∴4e2+4(e′)2=1 ⇒1e2+1(e′)2=14
Let line equation having intercepts as e,e′ be xe+ye′=1
since it is the tangent to the circle x2+y2=r2, so perpendicular distance from origin to the tangent is the radius of the circle. ∴∣∣
∣
∣
∣
∣
∣∣1√1e2+1(e′)2∣∣
∣
∣
∣
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∣∣=r ⇒r=2 units