A) 1
B) 4
C) 2
D) None of these
Correct Answer: C
Solution :
Since, tangents drawn from\[(1,2,\sqrt{3})\]to the ellipse\[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\]are right angled, therefore will lie on the > direction circle of the ellipse. So, \[(1,2,\sqrt{3})\]will lie on \[{{x}^{2}}+{{y}^{2}}=9+{{b}^{2}}\]. \[\therefore \] \[1+12=9+{{b}^{2}}\] \[\Rightarrow \] \[{{b}^{2}}=4\] \[\Rightarrow \] \[b=2\]You need to login to perform this action.
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