Parabola Equation Opening Downwards In The First Quadrant:

According to the definition of parabola, distance from focus F to point S is equal to distance from point S to the point T on directrix

➔    |FS| = |ST|

➔    [FS]² = [ST]²

➔    (x-h)²+(y-(k-p))² = (y-(k+p))2

➔    x²+h²-2xh+y²-2yk+2yp+k²+p²-2kp-2y(k-p) = y²+k²+p²+2pk-2yk-2yp

➔    Cancelling the matching terms ➔    x²+h²-2xh+y²-2yk+2yp+k²+p²-2kp-2yk = y²+k²+p²+2pk-2yk-2yp

➔    x²+h²-2xh+2yp-2kp = 2kp-2yp

➔    x²+h²-2xh+2yp-2kp-2kp+2yp = 0

➔    x²+h²-2xh-4kp+4yp = 0

➔    (x-h)²+4p(y-k) = 0

➔    (x-h)² = -4p(y-k)

When the vertex is at (0,0):

➔    h=0, k=0

➔    (x-h)² = -4p(y-k)

Parabola equation becomes: x² = -4py

Comparing the vertex form to the standard form :

➔    Standard form of parabola = a(x)²+b(x)+c

Comparing it with the vertex form gives the vertex as:

Distance from vertex to focus - let us call it "p".

Solved Example :