An equation that represents all points in the X-Y plane, equidistant from the point (1,1) & the line x = -1 ?

Question

a. y = x^2 /4 - x/2 + 1/4

b. y = x^2 /8 - x/4 - 7/8

c. x = -y^2 /4 + y/2 + 7/4

d. y = -x^2 /4 + x/2 + 7/4

e. y = -x^2 /8 + x/4 + 23/8

f. x = y^2 /4 - y/2 + 1/4

g. x = -y^2 /8 + y/4 + 23/8

h. x = y^2 /8 - y/4 - 7/8

or is it none of these ?

Answer 1

The set of points which are equidistant from the point (1,1) and the line x = -1 is traslated as follows:

Let (x, y) to be such a point, then it has to satisfy the
following condition

sqrt((x - 1)^2 + (y - 1)^2) = |1 * x + 0 * y + 1|/sqrt(1^2 + 0^2), i.e.
sqrt((x - 1)^2 + (y - 1)^2) = |x + 1|.

LHS is from the distance between (1,1) and (x,y).
RHS is from the distance between the line x + 1 = 0 and (x,y).

We square the both sides to obtain

(x - 1)^2 + (y - 1)^2 = |x + 1|^2, i.e.
(x - 1)^2 + (y - 1)^2 = (x + 1)^2, i.e.
x^2 - 2x + 1 + (y - 1)^2 = x^2 + 2x + 1, i.e.
(y - 1)^2 = 4x, i.e.
x = (1/4)(y - 1)^2

Seems like (f) is the answer, and it describes the parabola
with the vertex at (0,1).

Answer 2

This is the description of a parabola that opens sideways. The focus is (1,1) and the directrix is x = -1. The vertex is in the middle between the two. Its coordinates (h,k) are:

(h,k) = [(1 - 1)/2, 1) = (0,1)

The directed distance from the vertex to the focus is p.

p = 1 - 0 = 1
4p = 1

The equation of the parabola is of the form:

4p(x - 0) = (y - 1)²

4x = (y - 1)²

x = (1/4)(y - 1)²

x = y²/4 - y/2 + 1/4

The answer is f.

Answer 3

i get x = [(y-1)^2]/4
=y^2 /4 - y/2 - 1/4
the answer is (f)