(*) (7x2−4xy+8y2)2+(4x−8y−32) (7x2−4xy+8y2)=0 ⇒(7x2−4xy+8y2)(7x2−4xy+8y2+4x−8y−32) =0 Now, 7x2−4xy+8y2+4x−8y−32=0 ⇒7x2+(−4y+4)x+8y2−8y−32=0 Which is quadratic equationin x ∴x=
(4y−4)±√(16y2+16−32y)−4×7
×(8y2−8y−32)
2×7⇒x=
(2y−2)±√4y2+4−32y−56y2+56y+224
7
⇒x=
(2y−2)±√−52y2+24y+228
7
=
(2y−2)±√−4(13y2−6y−57)
7
Here, D<0 So, lines are not possible. Therefore, question is wrong.