A) \[\frac{{{x}^{2}}}{16}-\frac{{{y}^{2}}}{18}=1\]
B) \[\frac{{{x}^{2}}}{36}-\frac{{{y}^{2}}}{27}=1\]
C) \[\frac{{{x}^{2}}}{64}-\frac{{{y}^{2}}}{36}=1\]
D) \[\frac{{{x}^{2}}}{36}-\frac{{{y}^{2}}}{64}=1\]
E) \[\frac{{{x}^{2}}}{16}-\frac{{{y}^{2}}}{9}=1\]
Correct Answer: C
Solution :
\[\because \frac{2{{b}^{2}}}{{{a}^{2}}}=9\] Þ \[2{{b}^{2}}=9a\] ?..(i) Now \[{{b}^{2}}={{a}^{2}}({{e}^{2}}-1)=\frac{9}{16}{{a}^{2}}\]Þ\[a=\frac{4}{3}b\]?..(ii), (\[\because \]\[e=\frac{5}{4}\]) From (i) and (ii), \[b=6\], \[a=8\] Hence, equation of hyperbola\[\frac{{{x}^{2}}}{64}-\frac{{{y}^{2}}}{36}=1\].
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