Take three points a (1,2), B (m, n) C (4,4) on the parabola y2 = 4x whose focus is F. if the lengths FA, FB and FC of the three focal radii form an arithmetic sequence in turn What are the coordinates of point B?

Take three points a (1,2), B (m, n) C (4,4) on the parabola y2 = 4x whose focus is F. if the lengths FA, FB and FC of the three focal radii form an arithmetic sequence in turn What are the coordinates of point B?


analysis
2p=4
p=2
p/2=1
So focus f (0 1)
FA=(1 1)
FB=(M N-1)
FC=(4 3)
2FB=FA+FC
(2M 2(N-1))=(5 4)
therefore
2m=5
m=5/2
n-1=2
n=3
B(5/2 3)



P1 is the golden section point of line AB (AP1 is greater than BP1) O is the midpoint of AB and P2 is the symmetric point of P1 about O. to prove the median of P1B, P2b and p1p2


It is proved that: O is the midpoint, P2 is the symmetric point of P1 about o ∵ OP1 = op2, Ao = Bo ∵ ao-op2 = bo-op1, that is ap2 = BP1 and P1 is the golden section point of ab ∵ AP1 ^ 2 = BP1 * AB (ap2 + p1p2) ^ 2 = BP1 * (ap2 + p1p2 + BP1) (BP1 + p1p2) ^ 2 = BP1 * (2bp1 + p1p2) BP1 ^ 2 + p1p2 + 2bp1 * p1p2 = 2bp1 ^ 2 + B



Let p be a point on the hyperbola (x ^ 2 of 16) - (y ^ 2 of 9) = 1, and the distance from P to one focus of the hyperbola is 10, then what is the distance from P to another focus?
Detailed answers!


x^2/16-y^2/9=1,c=5,
The focus coordinates F1 (- 5,0), F2 (5,0), and,
a=4,
|PF1|=10,
|PF1|-|PF2|=2a,
|PF2|=10-8=2.
Or | PF2 | - | Pf1 | = 2A,
|PF2|=10+8=18,
The distance from P to another focus is 2, or 18, because according to the definition of hyperbola, the difference between the distance from two fixed points is hyperbola, which is the focus, and the difference is the distance between the real axis vertices (2a)



Take the right focus F2 that passes through the square of 16 / X-9 / y of hyperbola = 1 as the vertical line of X axis, and find the distance from the intersection m of this vertical line and hyperbola to the left focus F1


Set the focus to P
According to the first property of hyperbola, pf1-pf2 = 2A = 8
PF2 = B ^ 2 / a = 9 / 4
PF1=8-9/4=23/4



Given that the distance from a point P on the hyperbola x ^ 2 / 9-y ^ 2 / 16 = 1 to the left focus is 10, then the distance from P to the right focus is 10___


x^2/9-y^2/16=1
a=3,b=4
The distance from one point P to the left focus is 10
The distance to the right focus is equal to
10±2a=10±6
So it's 16 or 4



If the distance between a point on hyperbola x ^ 2 / 16-y ^ 2 / 18 = 1 and one focus is 12, what is the distance between a point and another focus


In hyperbola, a = 4
Let F1 and F2 be the two focuses,
From the definition of hyperbola, we can get | AF1 | - | af2 | = 2A = 8, if | af2 | = 12
There are | AF1 | - 12 | = 8, | AF1 | - 12 = ± 8, | AF1 | = 20 or 4
So the distance to the other focus is 20 or 4



Let P (6, m) be the point on the square of hyperbola X / 9-y / 16 = 1, and find the distance from P to the right focus of hyperbola


x^2/9-y^2/16=1.
a^2=9,b^2=16,c^2=9+16=25,c=5
That is, the coordinates of the right focus are (5,0)
P (6, m): 36 / 9-m ^ 2 / 16 = 1
The result is: m ^ 2 = 48
Therefore, the distance from P to the right focus = radical [(6-5) ^ 2 + (M-0) ^ 2] = radical (1 + m ^ 2) = radical (1 + 48) = 7



What is the distance between the focus of hyperbola 4 / x square = 1 and its asymptote?


Hyperbolic equation is incomplete, can't help you



If the distance between the point P on the hyperbola X & # 178 / 64-y & # 178 / 36 = 1 and its left focus is 8, then the distance between the point P and its right guide line is 8___________ The distance to its left collimator_______________


a=8
2a=16
c²=64+36=100
c=10
e=c/a=10/8=5/4
The distance from P to its left focus is 8



The distance from a point P on the hyperbola X & sup2 / 16-y & sup2 / 9 = 1 to its right collimator is 48 / 5, and the distance from P to the left focus is?
Detailed process


a=4
b=3
c=5
The right guide line is x = a ^ 2 / C = 16 / 5
If the point is on the right, then the distance from P to the right focus is 4
Then the left focus is 4 + 2 * 4 = 12
If the point is on the left, then the distance from P to the left collimator is 48 / 5 - 2 * 16 / 5 = 16 / 5
So the distance from P to the left focus is 4