Question

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

Answer 1

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

Explanation:

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

Answer 2

`x^2+y^2+xy-4y=0`, with vertex at V(3/4, 1/4) and size `a = sqrt 5/4`.

Explanation:

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`