How do you find the equation of the parabola having its focus at #( 1/2, 1/2)# and the directrix along #x - y = 1#?

Calling the directrix as

#L_d->p = p_0+lambda_d vec v_d# with
#p_0 = (0,-1)#, #vec v_d = (1,-1)# and #lambda_d in RR#

and the symmetry line

#L_s->p=p_f+lambda_s vec v_s# with
#p_f = (1/2,1/2)#, #vec v_s = (1,1)# and #lambda_s in RR#

we have #O = L_d nn L_s# computed as

#p_0+lambda_d vec v_d = p_f + lambda_s vec v_s#.

Solving for #lambda_d,lambda_s# we obtain #lambda_d=-1/2,lambda_s=-1 # and consequently
#O = (-1/2,-1/2)#

Now, the parabola is the place where the distance between a generic plane point #q=(x,y)# and #L_d# is equal to #norm(q-p_f)#.

but

#min norm(q-L_d) = sqrt(norm(q-O)^2- << q-O, hat(v_d) >>^2)#

where #hat(v_d) = vec v_d/norm(vec v_d)# so the parabola is the place of points #q# such that

#norm(q-O)^2- << q-O, hat(v_d) >>^2 = norm(q-p_f)^2#

giving

#x^2-4 x - 4 y - 2 x y + y^2 = 0#

Attached the parabola plot

Let (x, y) be a point on the parabola.

Using that its distance from the focus equals the distance from the

directrix,

#sqrt((x-1/2)^2+(y-1/2)^2))=+-(x-y-1)/sqrt(1^2+(-1)^2)#, two halves of the

parabola. Squaring,

#x^2+y^2-x-y+1/2=(1/2)(x^2+y^2+1-2xy-2x+2y)#. Upon simplification,

#x^2+y^2+xy-4y=0#

The vertex V of the parabola is on the perpendicular SX from the

focus S(1/2, 1/2), on the directrix DX given by x-y=1. The equation of

the perpendicular is of the form x+y=c. As S(1/2, 1/2) lies on this, c = 1.

#x-y=1 and x+y=1# meet at X( 1, 0).

V bisects SX.

So, V is (3/4, 1/4).. The size of the parabola

a = VS = #sqrt((3/4-1/2)^2+(1/4-1/2)^2)=sqrt(1/16+1/4)=sqrt 5/4#