Let each of the equations x^{2}+2xy+ay^{2}=0 | vedclass

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(B) Let $m = \frac{x}{y}$. The equations become $m^{2}+2m+a=0$ and $am^{2}+2m+1=0$. Since they have a common line, they share a common root $m$. Using

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$x+3y=0, 3x+y=0$

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(B) Let $m = \frac{x}{y}$. The equations become $m^{2}+2m+a=0$ and $am^{2}+2m+1=0$. Since they have a common line, they share a common root $m$. Using the condition for a common root: $(a_{1}b_{2}-a_{2}b_{1})(b_{1}c_{2}-b_{2}c_{1}) = (a_{1}c_{2}-a_{2}c_{1})^{2}$. Here, $(1 \cdot 2 - a \cdot 2)(2 \cdot 1 - a \cdot 2) = (1 \cdot 1 - a \cdot a)^{2}$. $2(1-a) \cdot 2(1-a) = (1-a^{2})^{2} \Rightarrow 4(1-a)^{2} = (1-a)^{2}(1+a)^{2}$. Since $a 
eq 1$, we have $4 = (1+a)^{2} \Rightarrow 1+a = \pm 2$. If $1+a=2$, $a=1$ (rejected as lines would be identical). If $1+a=-2$, $a=-3$. For $a=-3$, the equations are $x^{2}+2xy-3y^{2}=0$ and $-3x^{2}+2xy+y^{2}=0$. Factoring $x^{2}+2xy-3y^{2} = (x+3y)(x-y)=0$. Factoring $-3x^{2}+2xy+y^{2} = -(3x+y)(x-y)=0$. The common line is $x-y=0$. The other two lines are $x+3y=0$ and $3x+y=0$.

Let each of the equations $x^{2}+2xy+ay^{2}=0$ and $ax^{2}+2xy+y^{2}=0$ represent two straight lines passing through the origin. If they have a common line, then the other two lines are given by:

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