The equation $\frac{x^2}{12-k}+\frac{y^2}{8-k}=1$ represents hyperbola, if
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The equation $\frac{x^2}{12-k}+\frac{y^2}{8-k}=1$ represents hyperbola, if

(A) $k<8$

(B) $k>8$

(C) $8<k<12$

(D) None of these

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SOLUTION —

For represents a hyperbola,

$\begin{array}{lrlrl} & 12-k & >0 \\\text { and } & 8-k & <0 \\\Rightarrow & k & >12 \\\text { and } & k & >8 \\\therefore & 8 <k <12\end{array}$

So, The correct option will be (C).

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