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 :
Length of latus rectum \[=9=\frac{2{{b}^{2}}}{a}\] \[\Rightarrow \] \[{{b}^{2}}=\frac{9a}{2}\] ...(i) and \[e=\frac{5}{4}\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,1+\frac{{{b}^{2}}}{{{a}^{2}}}=\,\frac{25}{16}\] \[\therefore \] \[1+\frac{9a}{2{{a}^{2}}}=\frac{25}{16}\] [using Eq.(i)] \[\Rightarrow \] \[\frac{9}{2a}=\frac{9}{16}\] \[\Rightarrow \]\[a=8\] On putting this value in Eq. (i), we get \[{{b}^{2}}=\frac{9\times 8}{2}=36\Rightarrow b=6\] \[\therefore \]Equation of hyperbola is \[\frac{{{x}^{2}}}{{{8}^{2}}}-\frac{{{y}^{2}}}{{{6}^{2}}}=1\] \[\Rightarrow \] \[\frac{{{x}^{2}}}{64}-\frac{{{y}^{2}}}{36}=1\]
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