Try to find the parabolic equation of focus f (- 1,0) and quasilinear L: x = 3

Try to find the parabolic equation of focus f (- 1,0) and quasilinear L: x = 3

A:
Focus f (- 1,0), collimator x = 3
The symmetric line of x = - 1 and x = 3 is x = 1
So: the vertex of the parabola is at (1,0), and the opening is to the left
So: y ^ 2 = 2p (x-1), P

The Quasilinear of parabola is y = - x, the focus is [2,2]. What is the standard equation?

According to the characteristics of parabola, we get: (X-2) * (X-2) + (Y-2) * (Y-2) = {(x + y) / [the (1 / 2) power of 2]} square, the standard equation is: X * x + y * y-8x-8y + 16 = 2XY

It is known that the focal point of the parabola is (0,1 / 4) and the straight line L: y = x + 2. (1) find the standard equation of the parabola and its quasilinear equation
(2) Let the intersection of line L and parabola be a, B, and find the value of | ab |

1、
p/2=1/4
2p=1
The focus is above the origin and the opening is upward
x²=y
Guide line y = - 1 / 4
2、
y=x+2
x²=y=x+2
x²-x-2=0
x=2,x=-1
y=x+2
A(2,4),B(-1,1)
So AB = 3 √ 2

The Quasilinear equation of parabola y2 = - 8x is______ ．

∵ the equation of parabola is y2 = - 8x, ∵ 2p = 8, P = 4, ∵ its quasilinear equation is x = P2 = 2

Quasilinear equation of parabola y = 8x ^ 2

x²=y/8
That is 2p = 1 / 8
p/2=1/32
Opening up
So the guide line is below the x-axis
So y = - 1 / 32

The Quasilinear equation of parabola y2 = 8x is ()
A. x=-2B. x=-4C. y=-2D. y=-4

According to the parabolic equation, 2p = 8, P = 4, so the Quasilinear equation is x = - 2, so a is chosen

The Quasilinear equation of parabola y = - 1 / 8x & # 178; is
Why is the answer y = 2? It shouldn't be

In the parabola x2 = - 2PY, the focus is (0, - P / 2), and the equation of the Quasilinear is y = P / 2,
Y = - 1 / 8x & # 178; = - 8y
The equation of Quasilinear is y = 4 / 2 = 2

Solving quasilinear equation of parabola y ^ 2-8x + 6y + 17 = 0

By simplifying the original formula to (y + 3) ^ 2 = 8 (x-1), we can see that the image of the function is obtained by translating the function y ^ 2 = 8x. Specifically, the function image is translated one unit length to the right from X to X-1, and then y to y + 3. At this time, the function image is translated three unit lengths down

The Quasilinear equation of parabola y = 2px2 (P > 0) is______ ．

∵ x2 = 12py, the Quasilinear equation is y = - 18p, so the answer is: y = - 18p

It is known that the parabola C: y = 2px (P > 0) passes through point a (1, - 2). Find the equation of parabola C and its quasilinear equation
There is also a parabola of the intersection of the line L of the focus F of the parabola y = 2px (P > 0) at two points a (x1, Y1) B (X2, Y2). It is proved that x1x2 is a fixed value, and the value range of | ab | is obtained

Y = 2px (P > 0) passing point a (1, - 2)
(-2)^2=2p*1
p=2
y^2=2*2x=4x
Quasilinear equation x = - P / 2 = - 1
Passing parabola y ^ 2 = 2px (P > 0)
Focus coordinate f (P / 2,0)
Let K be the slope of the line
Y = K (X-P / 2), substituting y ^ 2 = 2px:
[k(x-p/2)]^2=2px
k^2x^2 - (k^2p+2p)x + k^2p^2/4 = 0
According to Weida's theorem: X 1x 2 = (k ^ 2p ^ 2 / 4) / K ^ 2 = P ^ 2 / 4 = fixed value, it is proved
When AB is parallel to y axis, AB is the shortest
x=p/2,y^2=2p*p/2 = p^2
AB = |y2-y1| = 2p
AB range [2p, + ∞)