If 4x^2+12xy+9y^2+2gx+2fy-1=0 represents a pa | vedclass

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(D) The general equation of a second-degree curve is $ax^2+2hxy+by^2+2gx+2fy+c=0$. For this to represent a pair of parallel lines,it must satisfy $h^2

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$\frac{f}{g}+\frac{g}{f}=\frac{13}{6}$

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(D) The general equation of a second-degree curve is $ax^2+2hxy+by^2+2gx+2fy+c=0$. For this to represent a pair of parallel lines,it must satisfy $h^2=ab$ and $af^2=bg^2$. Given $4x^2+12xy+9y^2+2gx+2fy-1=0$,we have $a=4, h=6, b=9$. Check condition $h^2=ab$: $6^2 = 36$ and $4 	imes 9 = 36$. This holds. Now use $af^2=bg^2$: $4f^2=9g^2$,which implies $f^2/g^2 = 9/4$,so $f/g = \pm 3/2$. Also,for parallel lines,the equation can be written as $(2x+3y+k_1)(2x+3y+k_2)=0$. Expanding this: $4x^2+12xy+9y^2+2k_1x+2k_2x+3k_1y+3k_2y+k_1k_2=0$. Comparing coefficients: $2g = 2(k_1+k_2) \implies g = k_1+k_2$ and $2f = 3(k_1+k_2) \implies 2f = 3g \implies f/g = 3/2$. Also $k_1k_2 = -1$. Since $g = k_1+k_2$ and $f = \frac{3}{2}g$,we have $f/g = 3/2$. Substituting into the options,$\frac{f}{g} + \frac{g}{f} = \frac{3}{2} + \frac{2}{3} = \frac{9+4}{6} = \frac{13}{6}$.

If $4x^2+12xy+9y^2+2gx+2fy-1=0$ represents a pair of parallel lines,then:

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