Let O(0, 0) and (0, 1) be two fixed points, then the locus of a point P such that the perimeter of `DeltaAOP` is 4, is

Options

A. `8x^(2)-9y^(2)+9y=18`
B. `9x^(2)-8y^(2)+8y=16`
C. `9x^(2)+8y^(2)-8y=16`
D. `8x^(2)+9y^(2)-9y=18`

Answer

C

Explanation

Given vertices of `DeltaAOP` are O(0, 0) and A(0, 1)
Let the coordinates of point P are (x, y)
Clearly, perimeter = OA + AP + OP = 4 (given)
`rArrsqrt((0-0)^(2)+(0- 1)^(2))+sqrt((0- x)^(2) + (1-y)^(2))+sqrt(x^(2) +y^(2))=4`
`rArr 1 + sqrt(x^(2)+(y-1)^(2))+sqrt(x^(2)+y^(2))=4`
`rArr sqrt(x^(2) + y^(2) - 2y +1) +sqrt(x^(2) + y^(2))=3`
`rArrx^(2)+y^(2)-2y + 1=3-sqrt(x^(2)+y^(2))`
`rArrx^(2)+y^(2)-2y +1=9+x^(2)+y^(2)-6sqrt(x^(2)+y^(2))`
[squaring both sides ]
`rArr 1- 2y = 9 - 6sqrt(x^(2)+y^(2))`
`rArr6sqrt(x^(2) + y^(2)) = 2y + 8 `
`rArr3sqrt(x^(2) + y^(2)) = y + 4 `
`rArr 9(x^(2) + y^(2)) = (y+4)^(2)` [ squaring both sides]
`rArr 9x^(2)+9y^(2)=y^(2)+8y + 16`
`rArr 9x^(2)+8y^(2)-8y=16`
Thus, the locus of point P(x,y) is
`9x^(2)+8y^(2)-8y = 16`