The straight line L passes through the focus of parabola C: y2 = 4x, and the slope k is greater than 2. The intersection of L and parabola C, a, B, AB, and the midpoint m to the straight line LM: 3x + 4Y + M = 0 (M > - 3) (- 2, - 3) distance is 1 / 5, then what is the value range of M?

The straight line L passes through the focus of parabola C: y2 = 4x, and the slope k is greater than 2. The intersection of L and parabola C, a, B, AB, and the midpoint m to the straight line LM: 3x + 4Y + M = 0 (M > - 3) (- 2, - 3) distance is 1 / 5, then what is the value range of M?

The focus is (1,0),
The equation of line L is
y=k(x-1)(k>2),
Substitution equation
k^2x^2-2(k^2+2)x+k^2=0.
Let the midpoint be (P, q)
p=(2/k^2)+1,q=2/k.
Where p ∈ (1,3 / 2), Q ∈ (0,1)
Substituting the formula of point line distance, we get
|3[(2/k^2)+1]+4(2/k)+m|=1.
Because of M > - 3, we have to
(6/k^2)+(8/k)+2+m=0,
∴m=-[(6/k^2)+(8/k)+2)∈(-15/2,-2),
So m ∈ (- 3, - 2)

It is known that the vertex of the parabola is at the origin, the Quasilinear equation x = 1, f is the focus, and the line passing through point a (- 2,0) intersects the parabola at P
(x1, Y1), q (X2, Y2), straight line PF, QF intersect parabola and point m, N, respectively,
(1) Solving the equation of parabola and the value of Y1 * Y2
(2) The slopes of PQ and Mn are K1 and K2 respectively

(1)
y^2=-4x
F(-1,0)
PQ:y=k*(x+2)
x=(y-2k)/k
y^2=-4y=-4*(y-2k)/k
ky^2+4y-8k=0
y1*y2=-8
(2)
k1=k(PQ)=(y1-y2)/(x1-x2)
The linear PF and QF are solved with the parabolic equation to obtain m (), n ()
k2=k(MN)=(yM-yN)/(xM-xN)

Through the focus F of parabola y ^ 2 = 2mx, make the vertical line of X axis intersect the parabola at two points a and B, and ab = 6
Solving the standard equation of parabola
I hope you can give me an answer before 10 o'clock,

Focus coordinates (M / 2,0)
Substituting M / 2 into the equation
y^2=m^2
y=+/- m
+m-(-m)=6
m=3
Y^2=6x

Find the standard equation of parabola suitable for the following conditions: 1. Make the vertical axis of X through the focus F of parabola y = 2mx (M is not equal to 0)
1. Through the focus F of parabola y = 2mx (M is not equal to 0), make the vertical line of X axis intersect the parabola at two points a and B, and / AB / = 6
2. The vertex of the parabola is at the origin, the axis of symmetry is the X axis, and the distance from the point P (- 5,2 times the root 5) to the focus is 6

1）
A(m/2,m),B(m/2,-m)
|AB|=±2m=6
m=±3
The standard equation of parabola: y ^ 2 = ± 3x
2)
The distance from point P (- 5,2 times the root 5) to the focus is 6
√[(p/2+5)^2+(2√5)^2]=6
(p/2+5)^2+20=36
P = - 2, or P = - 18
The standard equation of parabola: y ^ 2 = - 4x, or Y ^ 2 = - 36x

If the focus passing through the parabola y & # 178; = 4x is perpendicular to the X axis and intersects the parabola at two points a and B, then | ab|=

Parabola y & #178; = 4x
The focus is (1,0)
Let x = 1
We get y & # 178; = 4
So y = ± 2
So | ab | = 2 - (- 2) = 4
If you don't understand, I wish you a happy study!

Focus coordinates and standard equation of parabola with y ^ 2 = 6x-12

y^2=6x-12=6(x-2)
Let X-2 = x ', y = y' carry out coordinate translation
y'^2=6x'
The focus coordinate is (3 / 2,0)
Quasilinear equation x = - 3 / 2
Focus of Y ^ 2 = 6x-12 (7 / 2,0)
Quasilinear equation x = - 1 / 2

Finding the focal coordinate and quasilinear equation of parabola y 2 = 3x

y^2=3x
Focus coordinates: (3 / 4,0)
Quasilinear equation: x = - 3 / 4

Find a curve equation with focus f (3, - 3) and quasilinear y = 1

Take any point on the parabola, m (x, y). According to the definition of the parabola, the distance from m to f is equal to the distance to the directrix
That is: √ ((x-3) & sup2; + (y + 3) & sup2;) = | Y-1|
The square reduction is: (x-3) & sup2; = - 8 (y + 1)

It is known that the focal point of the parabola is (3,3) and the Quasilinear is the x-axis

We know that the lowest point of the parabola is point (3,1.5), so let the parabolic equation be (x-3) 2 = 2p (y-1.5). Because P / 2 = 1.5, so p = 3, so the parabolic equation is: (x-3) 2 = 6 (y-1.5), that is, y = 1 / 6 * (x-3) 2 + 1.5

To solve the Quasilinear equation of parabola y & #178; = X-1

This is obtained by translating y ^ 2 = x one unit to the right. The directrix of this parabola is x = - 1 / 4, and the directrix of the parabola in the question is x = 3 / 4