Has the directrix is vertical this is a horizontal parabola so use 4p(x − h) = (y − k)^2
where p is the distance from the directrix to the vertex along the axis of symmetry [and from the vertex to the focus] the direction matters as it give the sign of p. and (h,k) is the vertex
p = (3 − -2) = 5
=> (4*5)(x − 3) = (y − 1)^2
=> 20(x − 3) = (y − 1)^2
=> x − 3 = (1/20)(y − 1)^2
=> x = (1/20)(y − 1)^2 + 3 <== in vertex form.
=> x = (1/20)(y^2 − 2y + 1) + 3
=> 20x = y^2 − 2y + 61
=> y^2 − 20x − 2y + 61 = 0 <== in standard form.
where p is the distance from the directrix to the vertex along the axis of symmetry [and from the vertex to the focus] the direction matters as it give the sign of p. and (h,k) is the vertex
p = (3 − -2) = 5
=> (4*5)(x − 3) = (y − 1)^2
=> 20(x − 3) = (y − 1)^2
=> x − 3 = (1/20)(y − 1)^2
=> x = (1/20)(y − 1)^2 + 3 <== in vertex form.
=> x = (1/20)(y^2 − 2y + 1) + 3
=> 20x = y^2 − 2y + 61
=> y^2 − 20x − 2y + 61 = 0 <== in standard form.