If L_1=0 and L_2=0 are the asymptotes of the | vedclass

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(C) The equation of the given hyperbola is $9x^2-4y^2+36x+8y-4=0$. Rewriting the equation by completing the square: $9(x^2+4x+4) - 4(y^2-2y+1) = 4 + 3

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$\frac{81}{13}$

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(C) The equation of the given hyperbola is $9x^2-4y^2+36x+8y-4=0$. Rewriting the equation by completing the square: $9(x^2+4x+4) - 4(y^2-2y+1) = 4 + 36 - 4 = 36$. $9(x+2)^2 - 4(y-1)^2 = 36$. The combined equation of the asymptotes is obtained by setting the constant term to zero: $9(x+2)^2 - 4(y-1)^2 = 0$. This implies $3(x+2) = \pm 2(y-1)$. Thus,the two asymptotes are $L_1: 3x - 2y + 8 = 0$ and $L_2: 3x + 2y + 4 = 0$. The perpendicular distance from $(1,1)$ to $L_1$ is $d_1 = \frac{|3(1) - 2(1) + 8|}{\sqrt{3^2 + (-2)^2}} = \frac{9}{\sqrt{13}}$. The perpendicular distance from $(1,1)$ to $L_2$ is $d_2 = \frac{|3(1) + 2(1) + 4|}{\sqrt{3^2 + 2^2}} = \frac{9}{\sqrt{13}}$. The product of the distances is $d_1 	imes d_2 = \frac{9}{\sqrt{13}} 	imes \frac{9}{\sqrt{13}} = \frac{81}{13}$.

If $L_1=0$ and $L_2=0$ are the asymptotes of the hyperbola $9x^2-4y^2+36x+8y-4=0$,then the product of the perpendicular distances from the point $(1,1)$ to the lines $L_1=0$ and $L_2=0$ is

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