Question

#Quality Question

@Coordinate Geometry

Let O(0,0) and A(0,1) be two
fixed points,then the locus of point P such that the perimeter of △AOP is 4 , is ​

Answer 1

*Question:-

Let O(0,0) and A(0,1) be two

fixed points,then the locus of point P such that

the perimeter of ∆ AOP is 4 , is

*Answer:-

Given that,

OA + AP + OP = 4

=>$1 + \sqrt{ {x}^{2} + (y - {1)}^{2} } + \sqrt{ {x}^{2} + {y}^{2} } = 4$

$= > ( \sqrt{ {x}^{2} + {y}^{2} } {)}^{2} \\ = > (3 - \sqrt{ {x}^{2} + (y - 1 {)}^{2} {)}^{2} }$

${x}^{2} + {y}^{2} = 9 + {x}^{2} + {y}^{2} - 2y + 1 - 6 \sqrt{ {x}^{2} + (y - 1 {) }^{2} }$

$3 \sqrt{ {x}^{2} + {(y - 1)}^{2} } = (5 - y)$

$9( {x}^{2} + (y - 1 {)}^{2} ) = (5 - y {)}^{2}$

$= > {9x}^{2} + {8y}^{2} - 8y = 16$

Hope it helps you

@BrainlyTurtle

Answer 2

Answer:

Refer the attachment..