If the pair of straight lines 9x^2 + axy + 4y | vedclass

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(B) The general equation of a pair of lines is $Ax^2 + 2Hxy + By^2 + 2Gx + 2Fy + C = 0$. Comparing this with $9x^2 + axy + 4y^2 + 6x + by - 3 = 0$,we

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$a = 12, b = 4$

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(B) The general equation of a pair of lines is $Ax^2 + 2Hxy + By^2 + 2Gx + 2Fy + C = 0$. Comparing this with $9x^2 + axy + 4y^2 + 6x + by - 3 = 0$,we get $A=9, H=a/2, B=4, G=3, F=b/2, C=-3$.For parallel lines,$H^2 = AB$,so $(a/2)^2 = 9 	imes 4 = 36$,which gives $a^2 = 144$,so $a = \pm 12$.Also,for parallel lines,the ratio of coefficients of $x^2, xy, y^2$ must be the same as the ratio of coefficients of $x, y$ terms,i.e.,$A/G = H/F = G/C$ is not applicable here,but rather $A/H = H/B$ and $G/F = H/B$ or simply using the condition $h^2=ab$ and $af^2=bg^2$.Using $af^2 = bg^2$,we have $9(b/2)^2 = 4(3)^2$ $\Rightarrow 9(b^2/4) = 36$ $\Rightarrow b^2/4 = 4$ $\Rightarrow b^2 = 16$ $\Rightarrow b = \pm 4$.Testing $a=12, b=4$ in the equation $9x^2 + 12xy + 4y^2 + 6x + 4y - 3 = 0$,we get $(3x + 2y)^2 + 2(3x + 2y) - 3 = 0$. Let $t = 3x + 2y$,then $t^2 + 2t - 3 = 0 \Rightarrow (t+3)(t-1) = 0$. This represents two parallel lines $3x + 2y + 3 = 0$ and $3x + 2y - 1 = 0$.

If the pair of straight lines $9x^2 + axy + 4y^2 + 6x + by - 3 = 0$ represents two parallel lines,then:

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