If the square of the parabola y is equal to 2px (P is greater than 0), the distance between the big focus of the point with abscissa 2 is 3, then the value of P is 0

If the square of the parabola y is equal to 2px (P is greater than 0), the distance between the big focus of the point with abscissa 2 is 3, then the value of P is 0

Let the coordinates of that point be (x, y)
Because it's a parabola, then there's (y ^ 2 / 2p, y)
Because the abscissa is 2, that is, y ^ 2 / 2p = 2, y ^ 2 = 4P, y = ± 2 √ P
So the coordinates of that point are (2,2 √ P) or (2, - 2 √ P)
The focal coordinate of parabola is (P / 2,0)
Because the distance from that point to the focus is 3, then there is
(2-p/2)^2+(±2√p-0)^2=3^2=9
4-2p+p^2/4+4p=9
p^2/4+2p-5=0
p^2+8p-20=0
(p+10)(p-2)=0
Because P > 0, P = 2