A) \[-\frac{2}{\sqrt{3}}\]
B) \[\sqrt{3}\]
C) \[-\frac{\sqrt{3}}{2}\]
D) None of these
Correct Answer: A
Solution :
The equation of normal to the hyperbola \[\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{b}^{2}}}=1\]in terms of slope 'm' is \[y=mx\pm \frac{m({{a}^{2}}+{{b}^{2}})}{\sqrt{{{a}^{2}}-{{b}^{2}}{{m}^{2}}}}\] Given line \[y=mx+\frac{25\sqrt{3}}{3}\] and conic \[\frac{{{x}^{2}}}{16}-\frac{{{y}^{2}}}{9}=1\] which is hyperbola with \[{{a}^{2}}=16,\,{{b}^{2}}=9\] By comparing given line with equation of normal we get \[\pm \frac{m({{a}^{2}}+{{b}^{2}})}{\sqrt{{{a}^{2}}-{{b}^{2}}{{m}^{2}}}}=+\frac{25\sqrt{3}}{3}\] \[\Rightarrow \,\,\frac{m(16+9)}{\sqrt{16-9{{m}^{2}}}}=-\frac{25\sqrt{3}}{3}\] \[\Rightarrow \,\,\frac{25m}{\sqrt{16-9{{m}^{2}}}}=-\frac{25\sqrt{3}}{3}\] \[\Rightarrow \,9{{m}^{2}}=3(16-9{{m}^{2}})\] \[\Rightarrow {{m}^{2}}=\frac{16}{12}=\frac{4}{3}\Rightarrow \,m=\pm \frac{2}{\sqrt{3}}\]
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