(c) Given, $\frac{x^{2}}{36} - \frac{y^{2}}{k^{2}} = 1\ \Rightarrow \ k^{2} = \frac{36y^{2}}{x^{2} - 36}$ $k^{2} > 0\ if\ x^{2} - 36 > 0$ $\Rightarrow \ x^{2} > 36$ This is true only for point (10, 4). So, (10,4) lies on the hyperbola
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