If frac{x^{2}}{36}-frac{y^{2}}{k^{2}}=1 is a | vedclass

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(C) Given the equation of the hyperbola: $\frac{x^{2}}{36}-\frac{y^{2}}{k^{2}}=1$. Rearranging for $k^{2}$: $\frac{y^{2}}{k^{2}} = \frac{x^{2}}{36}-1

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$(10, 4)$ lies on the hyperbola

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(C) Given the equation of the hyperbola: $\frac{x^{2}}{36}-\frac{y^{2}}{k^{2}}=1$. Rearranging for $k^{2}$: $\frac{y^{2}}{k^{2}} = \frac{x^{2}}{36}-1 = \frac{x^{2}-36}{36}$ $k^{2} = \frac{36y^{2}}{x^{2}-36}$. Since $k^{2} > 0$,we must have $x^{2}-36 > 0$,which implies $x^{2} > 36$ or $|x| > 6$. Checking the options: $A$. $(-3, 1) \Rightarrow x^{2} = 9 $B$. $(3, 1) \Rightarrow x^{2} = 9 $C$. $(10, 4) \Rightarrow x^{2} = 100 > 36$ (True) $D$. $(5, 2) \Rightarrow x^{2} = 25 Therefore,the point $(10, 4)$ can lie on the hyperbola.

If $\frac{x^{2}}{36}-\frac{y^{2}}{k^{2}}=1$ is a hyperbola,then which of the following statements can be true?

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