If the distance between a point P on the hyperbola x2-y2 / 16 = 1 and one of its focal points is equal to 4, then the distance between point P and another focal point is equal to 4 Why does the answer fill in 6 have full marks and fill in 6 or 2 only have half marks? Please explain why 2 should be discarded

If the distance between a point P on the hyperbola x2-y2 / 16 = 1 and one of its focal points is equal to 4, then the distance between point P and another focal point is equal to 4 Why does the answer fill in 6 have full marks and fill in 6 or 2 only have half marks? Please explain why 2 should be discarded

A:
Hyperbola is the locus of a point whose absolute value of the distance difference between two fixed points on the plane is a fixed value
Hyperbola x ^ 2 - (y ^ 2) / 16 = 1
A ^ 2 = 1, B ^ 2 = 16, the focus is on the x-axis, a = 1
|PF1-PF2|=2a=2
PF2=4
Then:
|PF1-4|=2
Pf1-4 = 2 or pf1-4 = - 2
All:
Pf1 = 6 or Pf1 = 2
So the distance from point P to another focus is 2 or 6

If the ratio of the distance from a vertex of hyperbola to the corresponding quasilinear and the distance from this vertex to another focus is λ, then the value range of λ is

λ=(a-a^2/c)/(a+c)
=(AC-A ^ 2) / (AC + C ^ 2) (numerator and denominator divided by a ^ 2)
=(e-1)/(e+e^2),
1/λ=(e+e^2)/(e-1)=(e-1)+2/(e-1)+3
≥3+2√2,
So 0

What is the relationship between the distance from a point on the hyperbola to the focus and the distance to the Quasilinear

The distance from a point on the hyperbola to the focus is greater than the distance from the point on the hyperbola to the corresponding quasilinear
Equal to eccentricity e = C / A
That is, the distance from m to the left focus / the distance from m to the left collimator d = C / A
That is, the distance from m to the focal point / the distance from m to the right guide line d = C / A
This is the second definition of hyperbola

What is the relationship between the distance from a point on the hyperbola to the focus and the distance to the Quasilinear?

Let the standard equation of hyperbola be
x^2````y^2
----－----＝1(a＞0,b＞0)
a^2````b^2
If P is any point on the hyperbola, then
|PF|```c
----＝---
`d`````a

If | Pf1 |, | PF2 |, | F1F2 | is an arithmetic sequence with positive tolerance, the triangle can be obtained
It should be easy to understand

C^2=15+1=16 C=4
Because | Pf1 |, | PF2 |, | F1F2 | is an arithmetic sequence with positive tolerance
SO 2 | PF2 | = | Pf1 | + | F1F2 | = | Pf1 | + 8 (1)
Because | PF2 | - | Pf1 | = 2A = 2 (2)
PF2 = 6, Pf1 = 4
So the three sides are 4, 6 and 8
It should be

If the distance from a point m on the parabola y = 4x2 to the focus is 1, then the ordinate of point m is ()
A. 1716B. 1516C. 78D. 0

According to the definition of parabola, if the distance from m to the focus is 1, then the distance from m to the collimator is also 1. The collimator of ∵ parabola is y = - 116, and the ordinate of ∵ m point is 1-116 = 1516

The distance from a point m on the parabola f = 4x ^ 2 to the focus is 1. What is the ordinate of the point m

15/16
M point to the focus distance is equal to the distance to the guide line, calculate the ordinate of the guide line, and then 1 minus its absolute value on the line, understand

If the distance from a point m on the parabola y ^ 2 = 4x to the focus is 3, then the coordinates of the point m

y^2=4x
P = 2, the Quasilinear equation is x = - P / 2 = - 1
By definition, the distance from point m to focus is 3, which is equal to the distance from m to the guide line
Let the abscissa of M be x1
x1+1=3
x1=2
The abscissa is: 2
Y = ± 2 root sign 2
The coordinates of point m are (2, ± 2, root sign 2)

The distance from a point m to the focus on the parabola y ^ = 2px (P > 0) is a (a)

Because the definition of parabola is the set of points whose distance to a certain point is equal to that to a certain straight line, the distance to the directrix is a
Then your a should be > P / 2, because the minimum distance from the parabola to the focus is p / 2
Then the coordinate of this problem m should be (A-P / 2, + - radical [2p (A-P / 2)]

Given a (- 1,2), B (3,4), C (4, - 6), if the focus of the parabola y2 = ax is exactly the center of gravity of △ ABC, then a =
As shown in the picture

From the coordinate formula of the center of gravity, it is concluded that
x=(-1+3+4)/3=2,y=(2+4-6)/3=0
The center of gravity is (2,0)
Let the standard equation be y ^ 2 = 2px, then
p/2=2,p=4
a=2p=8.