Given x/a + y/b = 1 and ax + by = 1 are two variable lines, a,b being the parameters connected by relation a^2 + b^2 = ab. The locus of the point of the intersection has the equation lx^2 + my^2 + nxy - 1 = 0 then (l+m+n) is?

Question
Given x/a + y/b = 1 and ax + by = 1 are two variable lines, a,b being the parameters connected by relation a^2 + b^2 = ab. The locus of the point of the intersection has the equation lx^2 + my^2 + nxy - 1 = 0 then (l+m+n) is?

Answer
bx + ay = ab
and, ax + by = 1

Solving these equations,
x = a ( b^2 – 1 ) / ( b^2 – a^2 )
and, y = b ( a^2 – 1 ) / ( b^2 – a^2 )

Now,
x^2 + y^2
= ( 1 / ( b^2 – a^2 )^2 ) [ a^2 ( b4 – ^2 b^2 + 1 ) + b^2 ( a4 – ^2 a^2 + 1 ) ]
= ( 1 / ( b^2 – a^2 )^2 ) [ a^2 b4 – ^2 a^2 b^2 + a^2 + b^2 a4 – ^2 b^2 a^2 + b^2 ]
= ( 1 / ( b^2 – a^2 )^2 ) [ a^2 b4 + b^2 a4 – 4 a^2 b^2 + a^2 + b^2 ]
= ( 1 / ( b^2 – a^2 )^2 ) [ a^2 b^2 ( a^2 + b^2 ) – 4 a^2 b^2 + ( a^2 + b^2 ) ]

Since,
a^2 + b^2 = ab
x^2 + y^2
= ( 1 / ( b^2 – a^2 )^2 ) [ a^2 b^2 ( ab ) – 4 a^2 b^2 + ab ]
= ( ab / ( b^2 – a^2 )^2 ) [ a^2 b^2 – 4 ab + 1 ]
= ( ab / ( b^2 – a^2 )^2 ) [ a^2 b^2 – ab + 1 ] + ( ab / ( b^2 – a^2 )^2 ) [ – 3 ab ]
= ( ab / ( b^2 – a^2 )^2 ) [ a^2 b^2 – ab + 1 ] + ( ab / ( ( b^2 + a^2 )^2 – 4 a^2 b^2 ) ) [ – 3 ab ]
= ( ab / ( b^2 – a^2 )^2 ) [ a^2 b^2 – ab + 1 ] + ( ab / ( – 3 a^2 b^2 ) ) [ – 3 ab ]
= ( ab / ( b^2 – a^2 )^2 ) [ a^2 b^2 – ab + 1 ] + 1
= - xy + 1

Thus, x^2 + y^2 + xy – 1 = 0

Hence, (l+m+n) = 3.