A) 1
B) 2
C) 3
D) cannot be found
Correct Answer: B
Solution :
Since\[\frac{e}{2}\]and \[\frac{e'}{2}\]are eccentricities of hyperbola and its conjugate hyperbola hence,\[\frac{4}{{{e}^{2}}}+\frac{4}{e{{'}^{2}}}=1\] ?(1) equation of the line is \[\frac{x}{e}+\frac{y}{e'}=1\] \[e'x+ey-ee'=0\]which touches the circle \[{{x}^{2}}+{{y}^{2}}={{r}^{2}}\]Hence,\[\frac{ee'}{\sqrt{{{e}^{2}}+{{(e')}^{2}}}}=r\Rightarrow {{r}^{2}}=4\]or\[r=2\{\text{from}\,(1)\}\]You need to login to perform this action.
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