AP EAPCET 25-Aug-2021 Shift 2 Paper

Given line,
2x2−xy−y2=0 ...(i)
⇒ 2x2−2xy+xy−y2=0
2x(x−y)+y(x−y)=0
(2x+y)(x−y)=0
Since, we have given that combined equation of the pair is parallel to 2x2−xy−y2=0
So, combined equation will be
(2x+y+k1)(x−y+k2)=0 where k1 and k2 ...(ii)
are constants.
This equation satisfies (1,0).
∴ (2×1+0+k1)(1−0+k2)=0
⇒ k2=−1 and k1=−2
On putting the values of k1 and k2 in Eq. (ii)
(2x+y−2)(x−y−1)=0
⇒ 2x2−xy−y2−4x+y+2=0