A) \[4{{e}^{4}}-24{{e}^{2}}+35=0\]
B) \[4{{e}^{4}}+8{{e}^{2}}-35=0\]
C) \[4{{e}^{4}}-12{{e}^{2}}-27=0\]
D) \[4{{e}^{4}}-24{{e}^{2}}+27=0\]
Correct Answer: B
Solution :
Hyperbola is\[\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{b}^{2}}}=1\] \[\frac{a}{e}=\frac{4}{\sqrt{5}}\]and\[\frac{16}{{{a}^{2}}}-\frac{12}{{{b}^{2}}}=1\] \[{{a}^{2}}=\frac{16}{5}{{e}^{2}}\] ?.(1) and\[\frac{16}{{{a}^{2}}}-\frac{12}{{{a}^{2}}({{e}^{2}}-1)}=1\] ?(2) From (1) & (2) \[16\left( \frac{5}{16{{e}^{2}}} \right)-\frac{12}{({{e}^{2}}-1)}\left( \frac{5}{16{{e}^{2}}} \right)=1\]\[16\left( \frac{5}{16{{e}^{2}}} \right)-\frac{12}{({{e}^{2}}-1)}\left( \frac{5}{16{{e}^{2}}} \right)=1\] \[\Rightarrow 4{{e}^{4}}-24{{e}^{2}}+35=0\]You need to login to perform this action.
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