If the equation $\frac{x^2}{7-k}+\frac{y^2}{5-k}=1$ represents a hyperbola then
If the equation $\frac{x^2}{7-k}+\frac{y^2}{5-k}=1$ represents a hyperbola then
$5 \lt k \lt 7$
$k \lt 5$ or $k\gt7$
$k\gt5$
$k \neq 5, k \neq 7,-\infty \lt k \lt \infty$
Solution
$\mathrm{A} x^2+\mathrm{B} x y+\mathrm{C} y^2+\mathrm{D} x+\mathrm{E} y+\mathrm{F}=0$
represents a hyperbola if $B^2-4 A C\gt0, A \neq C$
Given equation : $\frac{x^2}{7-k}+\frac{y^2}{5-k}=1$
$\Rightarrow \mathrm{A}=\frac{1}{5-k}, \mathrm{~B}=0, \mathrm{C}=\frac{1}{7-k}$ So, $-4\left(\frac{1}{5-k}\right)\left(\frac{1}{7-k}\right)\gt0$
$4\left(\frac{1}{k-5}\right)\left(\frac{1}{k-7}\right) \lt 0$ $\therefore 5 \lt k \lt 7$
