Let `A3(0,4) and Bs(21,0) in R`. Let the perpendicular bisector of AB at M meet the y-axis at R. Then the locus of midpoint P of MR is `y=x^2 + 21`

Options

A. `x^(2) + y^(2) = (1)/(4) `
B. `(y-2)^(2) - x^(2) = 4 `
C. ` y + x^(2) = 2`
D. `3x^(2) + y^(2) = 8`

Answer

C

Explanation

The equation of MR is
`y - 2 = (t)/(2) (x - t) implies tx - 2y + 4 -t^(2) = 0`
It cuts y-axis at `( 0 , 2 - (t^(2))/(2))`
Let P , (h , k) be the mid-point of MR . Then ,
` h = (0 + t)/(2) , k = ( 2 - (t^(2))/(2) + 2)/(2)`
`implies t = 2h , k = 2 - (t^(2))/(4) implies k = 2 - h^(2) implies h^(2) + k = 2`
Hence , the locus of (h , k) is `x^(2) + y = 2`.