Point P is a point on the hyperbola x2 / 25-y2 / 9 with F1 and F2 as the focus, and the absolute value of Pf1 = 12, then the absolute value of PF2=

Point P is a point on the hyperbola x2 / 25-y2 / 9 with F1 and F2 as the focus, and the absolute value of Pf1 = 12, then the absolute value of PF2=

In hyperbola X & # 178 / 25-y & # 178 / 9 = 1, a = 5, then:
||Pf1 | - | PF2 | = 2A = 10, then:
|12 - | PF2 | = 10, get: | PF2 | = 2 or | PF2 | = 22

Given the hyperbola & nbsp; X29 − y216 = 1, F1 and F2 are its left and right focal points respectively, P is a point on the hyperbola, let | Pf1 | = 7, then the value of | PF2 | is______ ．

We know that a = 3 of hyperbola & nbsp; X29 − y216 = 1. From the definition of hyperbola, we know that | PF2 | - | Pf1 | = 2A = 6, | PF2 | - 7 = 6, | Pf1 | = 13

Find the standard equation of hyperbola with focus F 1 (0, - 6), F 2 (0,6) and passing through point m (2,5)

The focus is F1 (0, - 6), F2 (0,6),
C = 6, focus on y-axis
y^2/b^2-x^2/a^2=1
a^2+b^2=c^2
b^2=36-a^2
Put m in
25/(36-a^2)-4/a^2=1
25a^2-4(36-a^2)=a^2(36-a^2)
a^4-7a^2-144=0
(a^2-16)(a^2+9)=0
a^2>0
a^2=16
b^2=20
So y ^ 2 / 20-x ^ 2 / 16 = 1

It is known that the focus of hyperbola is F1 (- 6,0), F2 (6,0), and through P (- 5,2), the standard equation and quasilinear equation of hyperbola are obtained

c=6.
According to the definition, 2A = √ (11 & # 178; + 2 & # 178;) - √ (1 & # 178; + 2 & # 178;) = 5 √ 5 - √ 5 = 4 √ 5, a = 2 √ 5
So B = √ [6 & # 178; - (2 √ 5) &# 178;] = 4
So the standard equation is X & # 178 / 20-y & # 178 / 16 = 1
Alignment x = ± A & # / C = ± 10 / 3

The coordinates of the focal point are F1 (- 6,0), F2 (6,0) and pass through point a (- 5,2). The standard equation of the hyperbola is obtained

Let the equation be X & # 178; / A & # 178; - Y & # 178; / B & # 178; = 1, from the focus we know a & # 178; + B & # 178; = 6 & # 178; = 36, substituting point a to get 25 / A & # 178; - 4 / (36-a & # 178;) = 1, the solution A & # 178; = 20,
The equation is X & # 178 / 20-y & # 178 / 16 = 1

The focus F1 (- 4,0) F2 (4,0) of hyperbola is known, and the standard equation of hyperbola passing through point m (2 √ 6,2) is?

If the focus of hyperbola is F1 (- 4,0) F2 (4,0), then let the equation be x ^ 2 / A ^ 2-y ^ 2 / b ^ 2 = 1
M (2 radical 6,2) into: 24 / A ^ 2-4 / b ^ 2 = 1
And focus C = 4, that is, C ^ 2 = a ^ 2 + B ^ 2 = 16
The solution is: A ^ 2 = 12, B ^ 2 = 4
The equation is x ^ 2 / 12-y ^ 2 / 4 = 1

If the maximum distance from any point of an ellipse to its top vertex is exactly equal to the distance from the center of the ellipse to its directrix, the value range of eccentricity of the ellipse is obtained?
If B ^ 2 / C ^ 2 ≤ 1, the maximum value is a ^ 2 + B ^ 2 + B ^ 4 / C ^ 2 = a ^ 4 / C ^ 2
If B ^ 2 / C ^ 2 > 1, the maximum value is 4B ^ 2, which is equal to a ^ 4 / C ^ 2

Let the elliptic equation be x (x \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\+ (y-b) & 178; = A & 178; - A & 1

If the distance between the two focal points of the ellipse is equal to the distance between the endpoint of the major axis and the breakpoint of the minor axis, the eccentricity of the ellipse is calculated

The distance between two focal points of ellipse = 2C
The distance between the endpoint of the major axis and the breakpoint of the minor axis = root sign (a ^ 2 + B ^ 2)
So a ^ 2 + B ^ 2 = 4C ^ 2
2a^2-c^2=4c^2
2a^2=5c^2
c^2/a^2=2/5
So e = C / a = radical 10 / 5

If the eccentricity of ellipse 8 + m x + 9 y is 1 / 2, then the distance from the right focus to the left quasilinear is 0

x^2/(8+m)+y^2/9=1
c/a=1/2
c^2/a^2=1/4
a^2=8+m b^2=9
c^2=a^2-b^2=8+m-9=m-1
c^2/a^2=(m-1)/(m+8)=1/4
4m-4=m+8
3m=12
m=4
Ellipse: x ^ 2 / 12 + y ^ 2 / 9 = 1
C = radical (12-9) = radical 3
Guide line x = - A ^ 2 / C = - 12 / radical 3 = - 4 radical 3
Right focus (radical 3,0)
Distance from right focus to left collimator: radical 3 - (- 4 radical 3) = 5 radical 3

On the x-axis, the length of the real axis is 8, and an asymptote equation is the standard equation of a hyperbola with 3x-2y = 0

Let the standard equation of hyperbola be x ^ 2 / A ^ 2-y ^ 2 / b ^ 2 = 1,
Then 2A = 8, B / a = 3 / 2,
So a = 4, B = 6,
The standard equation of hyperbola is x ^ 2 / 16-y ^ 2 / 36 = 1