A) \[\frac{1}{10}\]
B) \[\frac{3}{10}\]
C) \[\frac{5}{10}\]
D) \[\frac{7}{10}\]
Correct Answer: B
Solution :
We have \[{{C}_{1}}:{{(x+1)}^{2}}+{{y}^{2}}=9\] \[{{C}_{2}}\,:{{(x-2)}^{2}}+{{y}^{2}}=49\] Now \[C{{C}_{1}}=r+{{r}_{1}}\] and \[C{{C}_{2}}\,={{r}_{2}}-r\] \[\Rightarrow \,\,C{{C}_{1}}+C{{C}_{2}}\,+{{r}_{1}}+{{r}_{2}}\] \[\therefore \] locus of C is an ellipse with focus at \[{{C}_{1}}\] and \[{{C}_{2}}\] Now \[{{r}_{1}}+{{r}_{2}}\,=2a=10\] ? and \[{{d}_{{{c}_{1}}{{c}_{2}}}}\] (focal length)\[=2ae=3\] ? \[\therefore \] and \[\Rightarrow \,\,\] eccentricity ?e? is \[\frac{3}{10}\]You need to login to perform this action.
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