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(D) The given equation is $9x^2-24xy+16y^2+\alpha x+\beta y+6=0$. This can be written as $(3x-4y)^2+\alpha x+\beta y+6=0$. Since it represents paralle

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$\frac{-3}{4}$

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(D) The given equation is $9x^2-24xy+16y^2+\alpha x+\beta y+6=0$. This can be written as $(3x-4y)^2+\alpha x+\beta y+6=0$. Since it represents parallel lines,let the lines be $(3x-4y+k_1)=0$ and $(3x-4y+k_2)=0$. Their product is $(3x-4y+k_1)(3x-4y+k_2) = 9x^2-24xy+16y^2+3(k_1+k_2)x-4(k_1+k_2)y+k_1k_2=0$. Comparing coefficients: $\alpha = 3(k_1+k_2)$,$\beta = -4(k_1+k_2)$,and $k_1k_2 = 6$. Thus,$\frac{\alpha}{\beta} = \frac{3(k_1+k_2)}{-4(k_1+k_2)} = -\frac{3}{4}$. The distance between parallel lines $Ax+By+C_1=0$ and $Ax+By+C_2=0$ is $\frac{|C_1-C_2|}{\sqrt{A^2+B^2}}$. Here,$\frac{|k_1-k_2|}{\sqrt{3^2+(-4)^2}} = 1 \implies |k_1-k_2| = 5$. Since $(k_1-k_2)^2 = (k_1+k_2)^2 - 4k_1k_2$,we have $25 = (k_1+k_2)^2 - 24$,so $(k_1+k_2)^2 = 49$,meaning $k_1+k_2 = \pm 7$. If one line passes through $(1,1)$,then $3(1)-4(1)+k = 0 \implies k = 1$. If $k_1=1$,then $k_2 = 6$ (since $k_1k_2=6$),so $k_1+k_2 = 7$. Then $\alpha = 3(7) = 21$ and $\beta = -4(7) = -28$. Thus,$\frac{\alpha}{\beta} = \frac{21}{-28} = -\frac{3}{4}$.

If $9x^2-24xy+16y^2+\alpha x+\beta y+6=0$ represents a pair of parallel lines $1$ unit apart and one of those lines passes through $(1,1)$,then $\frac{\alpha}{\beta} = $

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