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A. $\frac{x^2}{a^2}+\frac{y^2}{b^2}-z^2=k<0$ B.$\frac{x^2}{a^2}-\frac{y^2}{b^2}-\frac{z^2}{c^2}=1$ I don’t know how I can turn one into the other.

EDIT:Thanks to Bernard, I have somewhat of a clue as to where to begin. But Dividing the negative k on both both sides (assuming that the k is positive with a negative signs) and distribute the negative sign, I get $\frac{-x^2}{ka^2}-\frac{y^2}{kb^2}+\frac{z^2}{k}=1$ But this equation is still not identical with the second equation. Two of the three signs are different.

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Rewrite the first equation as $$\frac{z^2}{(\sqrt{-k})^2}-\frac{x^2}{(\sqrt{-k}\, a)^2}-\frac{y^2}{(\sqrt{-k}\, b)^2}=1.$$

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    Wouldn't the negative in the square root produce imaginary numbers?Dividing the negative k on both both sides (assuming that the k is positive with a negative signs) and distribute the negative sign, I get $\frac{-x^2}{ka^2}-\frac{y^2}{kb^2}+\frac{z^2}{k}=1$ But this equation is still not identical with the second equation. Two of the three signs are different. – 2017-01-16
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    If $k<0$ (that's the hypothesis you mentioned), $-k>0$. I just wanted to obtain a form as close to the second equation as possible. – 2017-01-16
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    But what does the difference in signs mean? Reversed along the z and y axes? – 2017-01-17
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    The *signature* of the quadratic form on the left is here $(1,2)$. It implies $z^2\ge\sqrt{-k}^2$, i.e. $z\ge\sqrt{-k}$ or $z\le-\sqrt{-k}$. This explains why `two` sheets. If the signature were $(2,1)$, it would imply $\frac{z^2}{(\sqrt{-k})^2}+\frac{x^2}{(\sqrt{-k}\, a)^2}\ge 1$, which explains why only `one` sheet. – 2017-01-17