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(C) The line perpendicular to the line $y=x$ has a slope of $-1$. Thus,the equation of the tangent line is $y = -x + c$ or $x + y - c = 0$.Given the h

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$3\sqrt{2}$

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(C) The line perpendicular to the line $y=x$ has a slope of $-1$. Thus,the equation of the tangent line is $y = -x + c$ or $x + y - c = 0$.Given the hyperbola equation $x^2 - 2y^2 = 18$,we rewrite it in standard form as $\frac{x^2}{18} - \frac{y^2}{9} = 1$.Here,$a^2 = 18$ and $b^2 = 9$.The condition for a line $y = mx + c$ to be a tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is $c^2 = a^2m^2 - b^2$.Substituting the values: $c^2 = 18(-1)^2 - 9 = 18 - 9 = 9$,so $c = \pm 3$.The equations of the tangent lines are $x + y + 3 = 0$ and $x + y - 3 = 0$.The distance $d$ between two parallel lines $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$ is given by $d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$.Substituting the values: $d = \frac{|3 - (-3)|}{\sqrt{1^2 + 1^2}} = \frac{6}{\sqrt{2}} = 3\sqrt{2}$ units.

The distance between the tangent lines to the hyperbola $x^2-2y^2=18$ which are perpendicular to the line $y=x$ is

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