A) \[\sqrt{3}\]
B) \[-\frac{2}{\sqrt{3}}\]
C) \[-\frac{\sqrt{3}}{2}\]
D) 1
Correct Answer: B
Solution :
We know that the equation of the normal of the conic \[\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{b}^{2}}}=1\] at point \[(a\sec \theta ,\,b\tan \theta )\] is \[ax\sec \theta +by\cot \theta ={{a}^{2}}+{{b}^{2}}\] or \[y=\frac{-a}{b}\sin \theta \,x+\frac{{{a}^{2}}+{{b}^{2}}}{b\cot \theta }\] Comparing above equation with equation \[y=mx+\frac{25\sqrt{3}}{3}\] and taking \[a=4,\,b=3\] we get, \[\frac{{{a}^{2}}+{{b}^{2}}}{b\cot \theta }=\frac{25\sqrt{3}}{3}\] \[\Rightarrow \]\[\tan \theta =\sqrt{3}\Rightarrow \theta ={{60}^{o}}\] and \[m=-\frac{a}{b}\sin \theta =\frac{-4}{3}\sin {{60}^{o}}\] = \[\frac{-4}{3}\times \frac{\sqrt{3}}{2}=\frac{-2}{\sqrt{3}}\].
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