A straight line passing through the focus of Y (2) = 2px (P > 0) intersects with this parabola. The ordinates of the two intersections are Y1 and Y2 respectively

A straight line passing through the focus of Y (2) = 2px (P > 0) intersects with this parabola. The ordinates of the two intersections are Y1 and Y2 respectively

Let the linear equation be y = K (X-P / 2) (K ≠ 0)
x=y/k+p/2
Substituting y ^ 2 = 2px, we get
y^2=2p(y/k+p/2)
y^2-(2p/k)y-p^2=0
So Y1 * y2 = - P ^ 2

Given the focus of line L passing through parabola y * 2 = 2px (P ＞ 0), and intersecting with parabola at two points (1) of a (x1, x2) and B (Y1, Y2), it is proved that y1y2 = - p * 2
(2) Point C is on the directrix of the parabola, and AC is parallel to the x-axis

Focus (P / 2,0) let Y / (X-P / 2) = 1 / N, x = NY + P / 2 be substituted into parabolic equation
Y ^ 2 = 2p (NY + P / 2) y ^ 2-2pny-p ^ 2 = 0 according to Weida's theorem; y1y2 = - P ^ 2, Y1 + y2 = 2pn
(2) Since C is on the guide line and AC is parallel to the X axis, AC is perpendicular to the guide line and C is perpendicular. Let f be the focus and AC = AF
Let a (x1, Y1) B (X2, Y2) f (+ P / 2,0) C (- P / 2, Y1) origin o (0,0). As long as it is proved that the slope of CO is equal to that of Bo, it is proved that CO and Bo are collinear. The slope of CO k = (Y1) / (- P / 2) Bo K ° = Y2 / X2, X2 = Ny2 + P / 2
k°-k=y2/(ny2+p/2) +2y1/p=[2py2 +2y1(2ny2+p)]/[(2ny2+p)*p]
=[2p(y1+y2)+4ny1y2]/[(2ny2+p)*p]
=[2p(2pn)+4n(--p^2)]/[(2ny2+p)*p]=[4np^2--4np^2]/[(2ny2+p)*p]=0
So B, C and the vertex of parabola are collinear

The distance from the focus to the collimator of the parabola y = 1 / 4x ^ 2 is equal to

The original equation can be reduced to x ^ 2 = 4Y
The focus is (0,1)
The Quasilinear equation is y = - 1
So the distance is two

If the distance between a point P and the y-axis on the parabola y2 = 4x is 3, then the distance between the point P and the focus F of the parabola is ()
A. 3B. 4C. 5D. 7

The Quasilinear equation of the parabola y2 = 4x is x = - 1 ∵ the distance from a point P on the parabola y2 = 4x to the y-axis is 3, the distance from P to the Quasilinear of the parabola is 4, and the distance from P to the focus F of the parabola is 4, so B is selected

It is known that the distance from a point P on the parabola y ^ 2 = 16x to the X axis is 12, and the focus is f|=

The parabolic focal coordinate is f (4,0), and the Quasilinear equation is x = - 4
Let P (x, y) and | y | = 12,
Because P is on a parabola, we have 12 ^ 2 = 16x, and the solution is x = 9,
|PF | is the distance from point P to the Quasilinear (the second definition of hyperbola), | PF | = 9 - (- 4) = 13

If the distance from the point with ordinate 6 on the parabola y = ax ^ 2 to the focus is 10, what is the distance from the focus to the directrix

y=ax^2
X ^ 2 = Y / A, the focal coordinate is (0,1 / (4a)), and the Quasilinear equation is y = - 1 / (4a)
If the distance from the point with ordinate 6 to the focus is 10, then the distance from the point to the guide line is also 10
That is: 6 + 1 / (4a) = 10
a=1/16.
So the focal coordinates are: (0,4) and the Quasilinear equation is x = - 4
Then the distance from the focus to the guide line is: 4 + 4 = 8

Given that the distance from the focus of the parabola y = AX2 to the collimator is 2, then the chord length of the parabola cut by the straight line y = x + 1 is equal to______ ．

Let the distance from the focus to the collimator of the parabola y = AX2 be 2, ∧ 12a = 2, ∧ a = 14 ∧ the parabola equation be y = 14x2, the focus is f (0,1), the collimator is y = - 1, ∧ the straight line y = x + 1 passes through the focus F, the straight line and the parabola equation are simultaneously established, and X is eliminated

Let the focus of the parabola c'y = 2px (P > 0) be f, let m be on C, | MF | = 5. If a circle with the diameter of MF passes through the point (0,2), then the equation of C is

F(0.5p,0)C:y^2=2pxM(2pa^2,2pa)(xM+xF)/2=0.25p+pa^2(yM+yF)/2=pa(2pa^2-0.5p)^2+(2pa)^2=MF^2=252pa^2+0.5p=5.(1)(0.25p+pa^2)^2+(pa-2)^2=(5/2)^2.(2)(1),(2):p=2,8,a=1,1/4C1:y^2=4xC2:y^2=16x

Let the parabolic standard equation c Focus f Point m is on C MF length is 5 If the circle with MF as diameter passes (0,2) Then the equation of C?

Let the focus of the parabola C: y2 = 3px (P ≥ 0) be f, the point m be on C, | MF | = 5. If a circle with the diameter of MF passes through the point (0,2), then the equation of C is

Take MF as diameter