Q.

If the pairs of lines x2+2xy+ay2=0 and ax2+2xy+y2=0 have exactly one line in common, then the join t equation of the other two lines is given by

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a

3x2+8xy−3y2=0

b

3x2+10xy+3y2=0

c

y2+2xy−3x2=0

d

x2+2xy−3y2=0

answer is B.

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Detailed Solution

Let y=mx be a line common to the given pairs of lines.Then, am2+2m+1=0 and m2+2m+a=0 ∴m22(1−a)=ma2−1=12(1−a)or m2=1 and m=−a+12 or (a+1)2=4or a=1or -3 But for a=1, the two pairs have both the lines common.So a=−3 and the slope m of the line common to both the pairs is 1.Now, ∴ x2+2xy+ay2=x2+2xy−3y2=(x−y)(x+3y) and ax2+2xy+y2=−3x2+2xy+y2=−(x−y)(3x+y) So, the equation of the required pair of lines is (x+3y)(3x+y)=0or 3x2+10xy+3y2=0
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