A) \[(-1-\sqrt{10},2)\]and \[(-1+\sqrt{10},2)\]
B) \[(-1-\sqrt{8},2)\]and \[(-1+\sqrt{8},2)\]
C) \[(-1,2-\sqrt{8})\]and \[(-1,2+\sqrt{8})\]
D) \[(-1,2-\sqrt{10})\]and \[(-1,2+\sqrt{10})\]
E) \[(\sqrt{10},0)\]and \[(-\sqrt{10},0)\]
Correct Answer: A
Solution :
Given equations can be rewritten as \[x+1=\sec t,\frac{y-2}{3}=\tan t\] Since, \[{{\sec }^{2}}t-{{\tan }^{2}}t=1\] \[\therefore \] \[\frac{{{(x+1)}^{2}}}{1}-\frac{{{(y-2)}^{2}}}{9}=1\] Now, \[e=\sqrt{1+\frac{{{b}^{2}}}{{{a}^{2}}}}=\sqrt{1+\frac{9}{1}}=\sqrt{10}\] \[\therefore \] \[Foci=(-1\pm ae,2)\] \[=(-1-\sqrt{10},2)\]and\[(-1+\sqrt{10},2)\]
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