The vertices of a hyperbola are $(0, 0)$ and $(10, 0)$ If one of its foci is $(18, 0)$ , the equation of the hyperbola is

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$\frac{x^{2}}{25} - \frac{y^{2}}{144} = 1$

$\frac{x^{2}}{25} - \frac{(y - 5)^{2}}{144} = 1$

$\frac{(x - 5)^{2}}{25} - \frac{(y - 5)^{2}}{144} = 1$

$\frac{(x - 5)^{2}}{25} - \frac{y^{2}}{144} = 1$

answer is B.

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Detailed Solution

Centre of the hyperbola is the mid point (5,0), of the vertices

Length of the transverse axis is $10 = 2 a \Rightarrow a = 5$ If $e$ is the eccentricity, then $a e - a = 18 - 10 = 8$

$\begin{array}{l} \Rightarrow e = 1 + \frac{8}{5} = \frac{13}{5} \\ b^{2} = a^{2} (e^{2} - 1) = 25 (\frac{169}{25} - 1) \\ = 144 \end{array}$

$\Rightarrow b = 12$ and the required equation of the hyperbola

is $\frac{(x - 5)^{2}}{25} - \frac{y^{2}}{144} = 1$

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