Solving the parabolic standard equation (1) the focus is f (- 7,0) (2) the collimator is y = 4 (3) the axis of symmetry is x, and the distance from the vertex to the focus is equal to 6

Solving the parabolic standard equation (1) the focus is f (- 7,0) (2) the collimator is y = 4 (3) the axis of symmetry is x, and the distance from the vertex to the focus is equal to 6

(1) The focus is f (- 7,0) P / 2 = 7, P = 14
The focus is on the x-axis, y ^ 2 = - 14x
(2) The guide line is y = 4 P / 2 = 8 2p = 32
Focus on Y-axis x ^ 2 = - 32x
(3) The axis of symmetry is x, and the distance from vertex to focus is equal to 6
p=6 2p=12
Y ^ 2 = 12x or Y ^ 2 = 12x

It is known that the vertex of the parabola is at the origin, the coordinate axis is the axis of symmetry, and through M (- 2, - 4), the equation of the parabola is solved

According to the meaning of the problem, let the parabola be y = ax ^ 2, and substitute the coordinates of point m into the above formula to get a = - 1, so the equation of the parabola is y = - x ^ 2

The equation of a circle whose center is on the parabola y2 = 2x (Y > 0) and tangent to the Quasilinear and x-axis of the parabola is ()
A. x2+y2−x−2y−14＝0B. x2+y2+x-2y+1=0C. x2+y2-x-2y+1=0D. x2+y2−x−2y+14＝0

Let the coordinates of the center of the circle be (B22, b), then B22 + 12 = B & nbsp; can be obtained from the tangent of the circle to the Quasilinear and x-axis of the parabola; so B = 1, so the center of the circle is (12, 1) radius r = 1, so the equation of the circle whose center is on the parabola y2 = 2x (Y > 0) and tangent to the Quasilinear and x-axis of the parabola is (x − 12) 2 + (Y − 1) 2 & nbsp; = 1, that is, X2 + Y2 − x − 2Y + 14 = 0, so D is selected