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# Given x/a+y/b=1 and ax+by=1 are two variable lines, 'a' and 'b' being the parameters connected by the relation a2+b2=ab. The locus of the point of intersection has the equation....A) x2+y2+xy-1=0B)x2+y2-xy+1=0C)x2+y2+xy+1=0D)x2+y2-xy-1=0

## 1 Answers

9 years ago

Dear Rounak,

bx + ay = ab

and, ax + by = 1

Solving these equations, x = a ( b2 – 1 ) / ( b2 – a2 )

and, y = b ( a2 – 1 ) / ( b2 – a2 )

Now, x2 + y2 = ( 1 / ( b2 – a2 )2 ) [ a2 ( b4 – 2 b2 + 1 ) + b2 ( a4 – 2 a2 + 1 ) ]

x2 + y2 = ( 1 / ( b2 – a2 )2 ) [ a2 b4 – 2 a2 b2 + a2  + b2 a4 – 2 b2 a2 + b2  ]

x2 + y2 = ( 1 / ( b2 – a2 )2 ) [ a2 b4 + b2 a4 – 4 a2 b + a2 + b2  ]

x2 + y2 = ( 1 / ( b2 – a2 )2 ) [ a2 b2 ( a2 + b2 ) – 4 a2 b + ( a2 + b2 ) ]

Since, a2 + b= ab

x2 + y2 = ( 1 / ( b2 – a2 )2 ) [ a2 b2 ( ab ) – 4 a2 b + ab ]

x2 + y2 = ( ab / ( b2 – a2 )2 ) [ a2 b2  – 4 ab  + 1 ]

x2 + y2 = ( ab / ( b2 – a2 )2 ) [ a2 b2  – ab  + 1 ] + ( ab / ( b2 – a2 )2 ) [  – 3 ab ]

x2 + y2 = ( ab / ( b2 – a2 )2 ) [ a2 b2  – ab  + 1 ] + ( ab / ( ( b2 + a2 )2 – 4 a2 b2 ) ) [  – 3 ab ]

x2 + y2 = ( ab / ( b2 – a2 )2 ) [ a2 b2  – ab  + 1 ] + ( ab / ( – 3 a2 b2 ) ) [  – 3 ab ]

x2 + y2 = ( ab / ( b2 – a2 )2 ) [ a2 b2  – ab  + 1 ] + 1

x2 + y2 = - xy + 1

Thus, x2 + y2 + xy – 1 = 0

Best Of luck

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