The equation frac{x^2}{1-k} - frac{y^2}{1+k} | vedclass

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(D) Given the equation $\frac{x^2}{1-k} - \frac{y^2}{1+k} = 1$ with $k > 1$.Since $k > 1$,we have $1-k  2$.Let $a^2 = k-1$ and $b^2 = k

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(D) Given the equation $\frac{x^2}{1-k} - \frac{y^2}{1+k} = 1$ with $k > 1$.Since $k > 1$,we have $1-k  2$.Let $a^2 = k-1$ and $b^2 = k+1$.Then the equation becomes $\frac{x^2}{-(k-1)} - \frac{y^2}{k+1} = 1$,which is $-\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.Multiplying by $-1$,we get $\frac{x^2}{a^2} + \frac{y^2}{b^2} = -1$.Since the sum of two squares cannot be negative,this equation represents an empty set (no real locus).However,if the question implies the standard form of a hyperbola where the signs of the denominators differ,we check the discriminant $B^2 - 4AC$.Here $A = \frac{1}{1-k}$ and $C = -\frac{1}{1+k}$.$B^2 - 4AC = 0 - 4(\frac{1}{1-k})(-\frac{1}{1+k}) = \frac{4}{1-k^2}$.Since $k > 1$,$k^2 > 1$,so $1-k^2 Thus,$B^2 - 4AC Given the structure,it does not represent a hyperbola for $k > 1$.

The equation $\frac{x^2}{1-k} - \frac{y^2}{1+k} = 1$,where $k > 1$,represents:

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