The center is at the origin, and a hyperbolic equation with focus (5,0) passing (4,0) is Please write the specific process

The center is at the origin, and a hyperbolic equation with focus (5,0) passing (4,0) is Please write the specific process

The focus is on the x-axis, let the equation be: X & sup2; / A & sup2; - Y & sup2; / B & sup2; = 1
Over (4,0), 16 / A & sup2; = 1, a = 4, B = 3
So the hyperbolic equation is: X & sup2 / 16-y & sup2 / 9 = 1

What is the trajectory equation of the center of gravity of hyperbolic focus triangle

Let the hyperbolic equation be: x ^ 2 / A ^ 2-y ^ 2 / b ^ 2 = 1, the focus triangle be: f1pf2, the center of gravity be g (XO, yo), P (x1, Y1) X1 ^ 2 / A ^ 2-y1 ^ 2 / b ^ 2 = 1 (*), then Po be (o coordinate origin) the middle line of the triangle, by the nature of the center of gravity, pg / go = 2, by the formula of fixed score point, XO = (1 / 3) x1, yo = (1 / 3) X2, X1 = 3xo, Y1 = 3

The hyperbola passes through a (- 2,4) B (4,4). If F1 (1,0) is one of its focal points, then the other focal point trajectory equation?

Solution: let the coordinates of the other focus be F2 (x, y), defined by hyperbola
||AF1|-|AF2||=||BF1|-|BF2||
That is | 5 - √ (x + 2) ^ 2 + (y-4) ^ 2 | = | 5 - √ (x-4) ^ 2 + (y-4) ^ 2|
∴√(x+2)^2+(y-4)^2=√(x-4)^2+(y-4)^2 ①
Or √ (x + 2) ^ 2 + (y-4) ^ 2 + √ (x-4) ^ 2 + (y-4) ^ 2 = 10
① The union of squares on both sides of the formula is reduced to x = 1
② The formula is √ (x + 2) ^ 2 + (y-4) ^ 2 = 10 - √ (x-4) ^ 2 + (y-4) ^ 2
Square union reduction, 5 √ (x-4) ^ 2 + (y-4) ^ 2 = 28-3x
Then, we can get [(x-1) ^ 2] / 25 + [(y-4) ^ 2] / 16 = 1
F2 does not coincide with F1
The locus of focus F 2 is straight line x = 1 and ellipse [(x-1) ^ 2] / 25 + [(y-4) ^ 2] / 16 = 1, but the point (1,0) should be excluded

Given that the hyperbola x ^ 2-y ^ 2 / 4 = 1 of the line L intersecting through the fixed point P (0,1) lies at two points a and B, Q: if the midpoint of the line AB is m, find the trajectory equation of point M
The line L passing through the fixed point P (0,1) is y = KX + 1
Substituting into hyperbola, we get 4x & # 178; - (KX + 1) &# 178; = 4, and finishing (4-K & # 178;) x & # 178; - 2kx-5 = 0 (1)
Let a and B be a (x1, Y1), B (X2, Y2) respectively, and the midpoint be m (x, y)
Then X1 + x2 = 2K / (4-K & # 178;), x1x2 = - 5 / (4-K & # 178;); Y1 + y2 = K (x1 + x2) + 2 = 2K & # 178; / (4-K & # 178;) + 2
For M: x = (x1 + x2) / 2 = K / (4-K & # 178;), y = (Y1 + Y2) / 2 = K & # 178; / (4-K & # 178;) + 1 = 4 / (4-K & # 178;)
Through the combination of algebraic expressions, we can get 4x & # 178; - Y & # 178; + y = 0
As for the range of Y: since there are two intersections, equation (1) must have two different roots
I really don't know how to draw it, but there are two ranges of drawing
This is the problem. How to combine the algebraic expressions?

x=k/(4-k^2) (1)
y=4/(4-k^2) (2)
From (2)
4-k^2=4/y
k^2=4-4/y
k=2√(1-1/y) (3)
Substituting (3) into (1)
x=2√(1-1/y)/(4-(4-4/y))
=2√(1-1/y)/(4/y)
=y√(1-1/y)/2
Square on both sides:
x^2=y^2/4*(1-1/y)=1/4*(y^2-y)
4x^2=y^2-y
4x^2-y^2+y=0

In equilateral △ ABC, Bo and co bisect ∠ ABC and ∠ ACB respectively, and the vertical bisectors of Bo and co intersect BC at e and f respectively? Why?

Be = CF, the reason is: connect OE, of, ∵ de vertical bisector ob ∵ be = OE (the distance from the point on the vertical bisector of the line segment to the point at both ends of the line segment is equal), similarly, of = CF, ∵ EBO = ∵ BOE, ∵ FCO = ∵ FOC, ∵ equilateral triangle ABC, ∵ ABC = ∵ ACB = 60 ° (the angles of the equilateral triangle are equal and 60 °)

In the triangle ABC, ab = AC, Bo and Co are bisectors of ∠ ABC and ∠ ACB respectively. Through O, EF is parallel to BC, indicating the relationship between BF + CE and Fe,

If EF ‖ BC and ob divide ∠ ABC equally,
∴∠FOB=∠OBC,∠FBO∠OBC,
It is found that ∠ FOB = ∠ FBO, ｜ FB = fo (1)
Similarly: EC = EO (2)
∴BF+CE=EF.

As shown in the figure, in △ ABC, the bisector of ∠ ACB intersects AB at point D, the parallel line passing through point D as BC intersects AC at point E, and the bisector of the outer angle of ∠ ACB intersects F

This paper: (1) the following: (1) the following: (1) the following: (1) the following: (1) the following: (1) (1) the "DCF = 12 ￣ ACB, 12 ACD = 12 ed = EC, EC =EF，∴DE=EF．

In the triangle ABC, the bisector of angle ACB intersects AB at e, the parallel line through e as BC intersects AC at F, and the bisector of angle ACD intersects g. it is proved that f is the midpoint of eg

Because eg ‖ BC
Therefore, GEC = BCE and EGC = DCG
Moreover, because ∠ ECF = ∠ BCE, ∠ GEC = ∠ ECF
So triangle ECF is isosceles triangle
So EF = CF
In the same way, it can be proved that the triangle CGF is isosceles triangle
So GF = CF
So EF = GF
So f is the midpoint of eg

As shown in the figure, in the triangle ABC, point O is a moving point on the side of AC. through point O, make a straight line Mn parallel to BC. The parallel line intersecting ∠ ACB is at point E and the outer angle intersecting ∠ ACB

1 & nbsp; proof: ∵ Mn / / BC
           ∴∠OEC=∠BCE   
          ∴∠OFC=∠FCG
&Nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; ∵ ∠ BCE = ∠ OCE (OE is the bisector of ∠ BCA)
           ∴∠OEC=∠OCE
            ∴OE=OC
&Nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; ∵ ∠ OCF = ∠ FCG (of is the outer bisector of ∠ BCA)
            ∴∠OCF=∠OFC
            ∴OF=OC
             ∴OE=OF
            ∴OC=1/2EF
When 2 & nbsp; & nbsp; o moves to the midpoint of AC side, the quadrilateral aecf is rectangular
&Proof: ∵ OE = OC
                    OE=OF
&When o is the midpoint of AC, OA = OC
          ∴OE=OC=OF=OA
The quadrilateral aecf is a rectangle

As shown in the figure, in the triangle ABC, the point m is the midpoint on the side of BC, the parallel line ab of bisector ad of BAC passing through point m is at F, and the extension line of intersection AC is at E
Verification: BF = CE

The extension line of BG ∥ AC to EM is made through B at g. ∥ BG ∥ AC, ∥ bgd ∥ CEM, ∥ GBM ∥ ECM, am = cm, ≌ BMG ≌ CNE, ≌ BG = CE. ∥ ad ∥ em, ∥ BFM = bad, ∥ CEM = CAD, and ∥ bad = CAD, ≌ BFM = CEM