Pair of straight lines

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If the pair of lines $x^{2} + 2 x y + a y^{2} = 0$ and $a x^{2} + 2 x y + y^{2} = 0$ have exactly one line in common, then $a =$

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Solution

Let $y = m x$ be a line common to the given pairs of
$a m^{2} + 2 m + 1 = 0$ and $m^{2} + 2 m + a = 0$
$\begin{array}{l} ∴ \frac{m^{2}}{2 (a - 1)} = \frac{m}{1 - a^{2}} = \frac{1}{2 (a - 1)} \\ \Rightarrow m^{2} = 1 and m = - \frac{a + 1}{2} \\ \Rightarrow (a + 1)^{2} = 4 \Rightarrow a^{2} + 2 a - 3 = 0 \Rightarrow a = - 3, 1 \end{array}$
But, for $a = 1,$ the two pairs have both the lines common.
Hence, $a = - 3 .$

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Pair of straight lines

Remember concepts with our Masterclasses.

If the pair of lines x2+2xy+ay2=0 and ax2+2xy+y2=0 have exactly one line in common, then a=

Let y=mx be a line common to the given pairs ofam2+2m+1=0 and m2+2m+a=0∴ m22(a−1)=m1−a2=12(a−1)⇒ m2=1 and m=−a+12⇒ (a+1)2=4⇒a2+2a−3=0⇒a=−3,1But, for a=1, the two pairs have both the lines common.Hence, a=-3.

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If the pair of lines $x^{2} + 2 x y + a y^{2} = 0$ and $a x^{2} + 2 x y + y^{2} = 0$ have exactly one line in common, then $a =$