Parabola y square = 20x find the Quasilinear equation and focus coordinates? Write the calculation process

Parabola y square = 20x find the Quasilinear equation and focus coordinates? Write the calculation process

Because the Quasilinear equation of parabola y square = 2px is x = - P / 2, and the focus is (P / 2,0)
So the Quasilinear equation of Y square = 20x is x = - 5, and the focus is (5,0)

(2014. Guang'an the first mock exam) the coordinates of the focal point of parabola x2=4y are ()
A. （1，0）B. （-1，0）C. （0，1）D. （0，-1）

∵ in the parabola x2 = 4Y, P = 2, P2 = 1, the focus is on the y-axis, the opening is upward, and the focus coordinate is (0, 1), so C

(2014. Guang'an the first mock exam) the coordinates of the focal point of parabola x2=4y are ()
A. （1，0）B. （-1，0）C. （0，1）D. （0，-1）

∵ in the parabola x2 = 4Y, P = 2, P2 = 1, the focus is on the y-axis, the opening is upward, and the focus coordinate is (0, 1), so C

If the intersection coordinate of the first-order function y = KX + B and X axis is (- 2,0), then the symmetry axis of the parabola y = ax & # 178; + BX is -

Straight line x = - 1