If the focus of y2 = 2x is f and the straight line passing through f intersects the parabola at two points AB, then the minimum value of | AF | + 4 | ab | Don't use the focal radius

If the focus of y2 = 2x is f and the straight line passing through f intersects the parabola at two points AB, then the minimum value of | AF | + 4 | ab | Don't use the focal radius

If the focal coordinate is (1 / 2,0), then let AB equation be x = my + 1 / 2
y^2=2(my+1/2),y^2-2my-1=0
y1+y2=2m,y1y2=-1
AB = radical (1 + m ^ 2) * | y1-y2 | = radical (1 + m ^ 2) * radical [4m ^ 2 + 4] = 2 (1 + m ^ 2)
AF=x1+p/2=x1+1/2.
When m = 0, the minimum value of AB is 2, then AB is perpendicular to X axis, X1 = 1 / 2, then AF = 1
The minimum value of AF + 4AB is 1 + 4 * 2 = 9

If there is a point m on the parabola y "= - 2px (P > 0), its abscissa - 9, its distance from the focus is 10
If there is a point m on the parabola y "= - 2px (P > 0), and its abscissa is - 9, its distance from the focus is 10. (1) find the parabolic equation. (2) find the distance from the focus of the parabola to the collimator

(1)y^2=-4x
(2)d=2

In the plane rectangular coordinate system xoy, the focus of the parabola y ^ 2 = 4x is f, the collimator is La, and B is the two moving points on the parabola ∠ AFB = 120 degree
If M is the midpoint of AB and M 'is the projection of m on L, then the maximum value of MM' / AB is obtained

Let AF = a, BF = B,
Defined by parabola, 2mm '= a + B
And the cosine theorem, AB2 = A2 + b2-2abcos120 ° = (a + b) 2-AB,
From a + B ≥ 2 √ (AB)
The results show that | ab | ≥ (√ 3 / 2) (a + b)
So the maximum value of MM '/ AB is √ 3 / 3

Given that the point P is on the parabola y2 = 4x, and the ratio of the distance from P to the Y axis to the distance from the focus is 12, then the distance from P to the X axis is ()
A. 14B. 12C. 1D. 2

Let the distance from P to the y-axis be a, then the distance from P to the focus is 2A. From the definition of parabola, we can get a + 1 = 2A, that is, the abscissa of P is 1. By substituting it into the parabolic equation, we can get the ordinate of P is ± 2, and the distance from P to the x-axis is 2

It is known that the square of (- 2,3) and parabola y = 2px, and the focal distance of P greater than 0 is 5. P=

From the problem, we know that the focus is (P / 2,0), the distance between it and (- 2,3) is (P / 2 + 2) ^ 2 + 3 ^ 2 = 5 under the root sign, and the square of both sides can be solved to get P = 4

Given that the parabolic equation is y2 = 2px (P > 0), the chord length of the straight line L: x + y = m passing through the focus of the parabola and cut by the parabola is 3, the value of P is obtained

The equation of line L is obtained as x + y = P2 by passing through the focus f (P2, 0) of parabola. By eliminating x + y = P2, y2 = 2px, Y2 + 2py-p2 = 0. From the meaning of the problem, it is obtained that △ = (2P) 2 + 4p2 > 0, Y1 + y2 = − 2p, y1y2 = − P2

If the straight line X-my + M = 0 passing through the focus of the parabola y ^ 2 = 2px (P > 0) intersects the parabola at two points a and B, and s △ AOB = 2 √ 2, then m ^ 6 + m ^ 4 =?

Connecting the linear equation with the parabolic equation, the solution is y ^ 2-2pmy + 2MP = 0 (Ya Yb) ^ = (Ya + Yb) ^ 2-4yayb = 4P ^ 2 m ^ 2-8mp. 1 / 2 * 1 * | Ya Yb | = 2 radical 2, and then the solution (pm-4) (PM + 2) = 0 (from P greater than 0, m greater than 0) is m = 4 / P. because the original line passes through the intersection point, the solution is 2P / 4 = m, P = 2m, so 2m ^ 2 = 4. M ^ 2 = 2

If one vertex of an equilateral triangle is at the focus of the parabola y & # 178; = 4x, and the other two vertices are on the parabola, then the side length of the equilateral triangle is

According to the characteristics: in the parabola y ^ 2 = 2px, the focus is (P / 2,0), we can see that the focus coordinates of the parabola y & # 178; = 4x are (1,0), let the other two vertex coordinates be (x1, Y1), (X2, Y2), because the triangle composed of these three points is an equilateral triangle, so there are X1 = X2, Y1 + y2 = 0, | Y1 | / | x1-1 | = tan30 ° =

The ratio of the distance from any point Q of parabola y ^ 2 = 2px (P > 0) to the vertex o to the distance from the focus f is k, and the value range of K is obtained

The parameter equation of parabola is x = 2pt ^ 2, y = 2pt parabola Q (2pt ^ 2,2pt) | OQ | = 2pt √ (T ^ 2 + 1) | QF | = 2pt ^ 2 + 1 / 2pk = (2pt √ (T ^ 2 + 1)) / (2pt ^ 2 + 1 / 2P) = 4T √ (T ^ 2 + 1) / (4T ^ 2 + 1) = 4 / √ 3 * (√ 3T) √ (T ^ 2 + 1) / ((4T ^ 2 + 1)) ≤ 2 / √ 3 × (3T ^ 2 + T ^ 2 + 1) / (4T ^

A straight line passing through the focus of the parabola y ^ 2 = 2px intersects the parabola, and the ordinates of the two focuses are Y1 and Y2. Verify that y1y2 = - P ^ 2

The focus of parabola y ^ 2 = 2px is (P / 2,0)
So let the linear equation with this focus be y = K (X-P / 2)
The parabola y ^ 2 = 2px and the straight line y = K (X-P / 2) can be obtained simultaneously
k^2(x-p/2)^2=2px
That is k ^ 2x ^ 2 - (k ^ 2-2) PX + K ^ 2p ^ 2 / 4 = 0
Therefore, the coordinates of the two intersections can be set as (x1, Y1), (X2, Y2) according to the meaning of the title
therefore
x1*x2=p^2/4
So (y1y2) ^ 2 = 2px1 * 2px2 = 4P ^ 2 * P ^ 2 / 4 = P ^ 4
Because Y1 and Y2 must be positive and negative (the straight line passing through the focal point must take the parabola intersecting the Y positive half axis and Y negative half axis respectively)
So y1y2 = - P ^ 2
*This is a multiplier sign
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