If one of the pair of lines 4x2+6xy+ky2=0 is perpendicular to one of die lines represented by 3x2−5xy+2y2=0, then twice the absolute difference of such possible values of k is
We have 4x2+6xy+ky2=0.....(i) and 3x2−5xy+2y2=0....(ii) Let y=mx be one of the lines represented by Eq. (i). Then, y=mx satisfies Eq. (i) Eq. (i). Then, y=mx satisfies Eq. (i) ∴4x2+6x⋅mx+km2n2=0 ⇒‌‌km2+6m+4=0.....(iii) The equation of a line passing through the origin and perpendicular to y=mx, is y−0=−‌
1
m
(x−0)⇒my+x=0 Since, one of the lines given by Eq. (i) is perpendicular to one of the lines given by 3x2−5xy+2y2=0. Therefore one of the lines given by (ii) is my+x=0 Hence, my+x=0 i.e., x=−my satisfies Eq. (ii). ∴‌‌3(−my)2−5y(−my)+2y2=0 ⇒‌‌3m2y2+5my2+2y2=0 ⇒‌‌3m2+5m+2=0....(iv) using Eqs. (iii) and (iv), we get