A) \[\sqrt{3}x-y+7=0\]
B) \[\sqrt{3}x-y-7=0\]
C) \[\sqrt{3}x-y\pm 7=0\]
D) None of these
Correct Answer: A
Solution :
Given, \[m=\tan 60{}^\circ =\sqrt{3}\] Also ellipse, \[{{x}^{2}}+16{{y}^{2}}=16\] \[\Rightarrow \] \[\frac{{{x}^{2}}}{16}+\frac{{{y}^{2}}}{1}=1\] Here, \[{{a}^{2}}=16,{{b}^{2}}=1\] \[\therefore \]Equation of tangent \[y=\sqrt{3}x\pm \sqrt{16.3+1}\] [From\[y=mx\pm \sqrt{{{a}^{2}}{{m}^{2}}+{{b}^{2}}}\]] \[\Rightarrow \] \[y=\sqrt{3}x\pm \sqrt{49}\] \[\Rightarrow \] \[y=\sqrt{3}x\pm 7\] Hence, required equation is\[\sqrt{3}x-y\pm 7=0\]
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