Find the equation of the parabola

a parabola satisfies the following conditions
vertex on x-y-2=0
focus on x+y-2=0
with vertical directrix; and passes through the point (9,7)
find the equation(s) of the parabola(s) write them in standard form

The equations are
(y-7)^2 = -56(x-9) and (y+1)^2 = 8(x-1).
Firstly, assume a general point on both the given lines; (a,a-2) and (b,2-b).
Now, since the directrix is vertical, a-2 = 2-b
or b = 4-a. Putting this value of b in the point,
we get vertex and focus as (a,a-2) and (4-a,a-2) respectively.
Now, distance between the focus and vertex = 2a-4 = distance between vertex and the directrix.
So, equation of the directrix is x = a+(2a-4) = 3a-4.
Now, using the property of parabola that for every point on parabola, its distance from focus is equal to that from directrix, we get
[{x-(4-a)}^2 + {y-(a-2)}^2]^(1/2) = x-(3a-4) ........ A
Put (9,7) in the equation and squaring both sides,
we obtain a quadratic equation in a,
a^2 - 10a + 9 = 0.
Solving this, we get a=1,9.
Putting these values of a in A,
we obtain two equations,
(y-7)^2 = -56(x-9) and (y+1)^2 = 8(x-1).