Question

Let A(-3,0)and B(3,0) are two fixed points. Find the equation of locus of a point P which moves so that AP:BP=1:2

Answer 1

Answer:

3x² + 3y² + 30x + 27 = 0

Step-by-step explanation:

Let P be ( x, y ). Here, AP : BP = 1 : 2

= > AP² : BP² = 1² : 2²

= > AP² : BP² = 1 : 4

= > 4AP² = BP²

Using distance formula -

= > AP = √[ ( - 3 - x )² + ( 0 - y )² ]

= > AP² = ( x + 3 )² + y²

= > BP = √[ ( 3 - x )² + ( 0 - y )² ]

= > BP² = ( 3 - x )² + y²

From above,

= > 4AP² = BP²

= > 4{ ( x + 3 )² + y² } = ( 3 - x )² + y²

= > 4( x² + 9 + 6x ) + 4y² = 9 + x² - 6x + y²

= > 4x² + 36 + 24x + 4y² = 9 + x² - 6x + y²

= > 3x² + 3y² + 30x + 27 = 0

Equation of that locus is 3x² + 3y² + 30x + 27 = 0