If the pairs of lines `x^(2)+2xy+ay^(2)=0andax^(2)+2xy+y^(2)=0` have exactly one line in common then the joint equation of the other two lines is given by

Options

A. `3x^2+8xy-3y^2=0`
B. `3x^2+10xy+3y^2=0`
C. `y^2+2xy-3x^2=0`
D. `x^2+2xy-3y^2=0`

Answer

B

Explanation

Let `y=mx` be a line common to the given pairs of lines. Then , `am^2+2m+1=0 and m^2+2m+a=0`
`rArr (m^2)/(2(a-1))=(m)/(a^2-1)=(1)/(2(a-1))`
`rArr m^2=1 and m=(a+1)/(2)`
`rArr (a+1)^2=4rArr a=1 or -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 boththe pairs is 1.
Now , `x^2+2xy+ay^2=x^2+2xy-3y^2=(x-y)(x+3y) and ax^2+2xy+y^2=-3x^2+2xy+y^2=-(x-y)(3x+y)`
So, the equation of the required lines is `(x+3y)(3x+y)=0`
i.e., `3x^2+10xy+3y^2=0`.