A) \[\frac{{{x}^{2}}}{25}-\frac{{{y}^{2}}}{144}=1\]
B) \[\frac{{{(x-5)}^{2}}}{25}-\frac{{{y}^{2}}}{144}=1\]
C) \[\frac{{{x}^{2}}}{25}-\frac{{{(y-5)}^{2}}}{144}=1\]
D) \[\frac{{{(x-5)}^{2}}}{25}-\frac{{{(y-5)}^{2}}}{144}=1\]
Correct Answer: B
Solution :
\[2a=10\], \ \[a=5\] \[ae-a=8\]or\[e=1+\frac{8}{5}=\frac{13}{5}\] \\[b=5\sqrt{\frac{{{13}^{2}}}{{{5}^{2}}}-1}=5\times \frac{12}{5}=12\] and centre of hyperbola\[\equiv (5,\,0)\] \\[\frac{{{(x-5)}^{2}}}{{{5}^{2}}}-\frac{{{(y-0)}^{2}}}{{{12}^{2}}}=1\].
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