A parabola is drawn with focus at (3,4) and vertex at the focus of the parabola `y^2-12y-4y+4=0`. The equation of the parabola is

Options

A. `y^(2)-8x-6y+25=0`
B. `y^(2)-6x-8y-25=0`
C. `x^(2)-6x-8y+25=0`
D. `x^(2)+6x-8y-25-0`

Answer

C

Explanation

Given equation can be rewritten as `(y-2)^(2)=12x` Here vertex and focus are (0,2) and (3,2).
`:.` Vertex of the required parabola is (3,2) and focus is (3,4). The axis of symmetry is `x=3` and latusrectum
`=4xx2=8`
Hence, required equation is
`(x-3)^(2)=8(y-2)`
`impliesx^(2)-6x-8y+25=0`