A) \[\frac{4}{3}\]
B) \[\frac{4}{\sqrt{3}}\]
C) \[\frac{2}{\sqrt{3}}\]
D) None of these
Correct Answer: C
Solution :
| [c] The standard equation of hyperbola is |
| \[\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{b}^{2}}}=1\] |
| Latus rectum \[=\frac{2{{b}^{2}}}{a},\] Conjugate axis\[=2b\], |
| Distance between the foci =2ae |
| According to the question, |
| \[\frac{2{{b}^{2}}}{a}=8\] ? (i) |
| \[2b=\frac{1}{2}(2ae)\Rightarrow b=\frac{ae}{2}\] ? (ii) |
| From (i) & (ii), \[\frac{2}{a}{{\left( \frac{ae}{2} \right)}^{2}}=8\] |
| \[\Rightarrow 2.\frac{{{a}^{2}}{{e}^{2}}}{4a}=8\] |
| \[\Rightarrow a{{e}^{2}}=16\] ? (iii) |
| From (i), \[{{b}^{2}}=4a\] Using \[{{b}^{2}}={{a}^{2}}({{e}^{2}}-1)\] we get |
| \[(4a)={{a}^{2}}({{e}^{2}}-1)\Rightarrow 4=\frac{16}{{{e}^{2}}}({{e}^{2}}-1)\] |
| \[\Rightarrow 16-\frac{16}{{{e}^{2}}}=4\Rightarrow \frac{16}{{{e}^{2}}}=12\therefore e=\frac{2}{\sqrt{3}}.\] |
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