If the equation x+y+n=0 represents a normal t | vedclass

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(B) The given equation of the line is $x+y+n=0$,which can be written as $y=-x-n$.Comparing this with $y=mx+c$,we get $m=-1$ and $c=-n$.The condition f

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$\pm 4$

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(B) The given equation of the line is $x+y+n=0$,which can be written as $y=-x-n$.Comparing this with $y=mx+c$,we get $m=-1$ and $c=-n$.The condition for the line $y=mx+c$ to be a normal to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is $c^2 = \frac{(a^2+b^2)^2 m^2}{a^2-b^2 m^2}$.Here,$a^2=6$ and $b^2=2$.Substituting the values,we get $(-n)^2 = \frac{(6+2)^2 (-1)^2}{6-2(-1)^2}$.$n^2 = \frac{8^2 	imes 1}{6-2} = \frac{64}{4} = 16$.Therefore,$n = \pm 4$.

If the equation $x+y+n=0$ represents a normal to the hyperbola $\frac{x^2}{6}-\frac{y^2}{2}=1$,then $n=$

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