A) \[(4\text{ }sec\theta ,2\text{ }tan\theta )\]
B) \[(4\text{ }sec\theta +1,\text{ }2\text{ }tan\theta -2)\]
C) \[(4\text{ }sec\theta -1,2\text{ }tan\theta -2)\]
D) \[(sec\theta -4,\text{ }tan\theta -2)\]
E) \[(4\text{ }sec\theta -1,\text{ }2\text{ }tan\theta +2)\]
Correct Answer: E
Solution :
Any point on the hyperbola \[\frac{{{(x+1)}^{2}}}{16}-\frac{{{(y-2)}^{2}}}{4}=1\]is of the form \[\{(4\sec \theta -1),(2\tan \theta +2)\}\]
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