If the absolute value of AB is 7, then the linear equation is Is there a simpler way

If the absolute value of AB is 7, then the linear equation is Is there a simpler way

By using the formula of focal radius, a + EX1 + A + ex2 = 7,2a + e (x1 + x2) = 7. Let the linear equation be a point oblique equation. Let the elliptic equation be simultaneous, and the value of K can be obtained by using Weida's theorem
As shown in Figure 1, if ad is the bisector of an internal angle of △ ABC, then ab ∶ AC = BD ∶ CD. This conclusion is the bisector theorem of an internal angle of a triangle
Attached figure
You draw your own picture
CE is parallel to Da through point C, and the extension of Ba is parallel to E
^E=^BAD=^DAC=^ACE SO AC=AE
Adiiec BA / AE = BD / DC uses AC instead of AE so AB / AC = BD / DC
The line L passing through the right focus of ellipse C: 3x ^ 2 + 4Y ^ 2 = 12 intersects ellipse C at two points ab. if the sum of the distances from two points AB to the right guide line is 7, the equation of line L is obtained
x^2/4+y^2/3=1
a=2 c=1
Guide line x = a ^ 2 / C = 4
L:y=k(x-1)
Bring in
3x^2+4y^2=12
3x^2+4[k(x-1)]^2=12
(3+4k^2)x^2-8k^2x+4k^2-12=0
The sum of the distances from two points AB to the right guide line is 7
4-x1 + 4-x2 = 7 (4 must be greater than x1, x2)
Using the great theorem
It can be solved
In triangle ABC, the angle BAC is equal to 120 ° and ad bisects the angle BAC.AB Is 5, AC is 3, ad?
Ad long!
Let the length of ad be x, then de = DF = 2 / 3 root X
2S = 3 * 5 * sin60 = 15 / 2 root sign 3
Then 5de + 3DF = 2S, x = 15 / 8
If a straight line passing through the intersection of the left focus of ellipse 3x ^ 2 + 4Y ^ 2 = 18 with slope of 1 intersects the ellipse at points a and B, how long is ab?
The elliptic equation is: x ^ 2 / 16 + y ^ 2 / 12 = 1,
a=4,b=2√3,c=2,
Centrifugation e = C / a = 1 / 2,
Using the point difference method,
Let a (x1, Y1), B (X2, Y2),
x1^2/16+y1^2/12=1.(1)
x2^2/16+y2^2/12=1.(2)
(1) (2) formula,
3/4+[(y1-y2)/(x1-x2)][(y1+y2)/2]/[(x1+x2)/2]=0
Where (y1-y2) / (x1-x2) is the slope of the line K,
(Y1 + Y2) / 2 and (x1 + x2) / 2 are the ordinates (- 1,1) and abscissa (- 1,1) of the midpoint of AB, respectively,
∴3/4+k*(-1)=0,
∴k=3/4,
Let the angle between AB and X be θ,
tanθ=k=3/4,
secθ=√（1+9/16）=5/4,
cosθ=4/5,
From the formula of chord length through focus:
|AB|=(2b^2/a)/[1-e^2(cosθ)^2]=(2*12/4)/[1-(1/2)^2*(4/5)^2]
=50/7.
∴|AB|=50/7.
As shown in the figure, in △ ABC, a = 2 ∠ B, CD is the bisector of ACB, if AC = 5, BD + BC = 18, then ab=______ .
Intercept CE = AC on BC and connect de. ≌ CD is bisector of ≌ ACB, ∵ ACD = ∠ ECD, ∵ in △ ACD and △ ECD, CE = AC ≌ ECD = CD, ≌ ACD ≌ ECD (SAS), ≌ ad = ed, AC = CE, ≂ a = CED, ≂ a = 2 ≂ B, ≂ CED = 2 ≌ B, ≌ ad = ed, AC = CE, ≌ a = CED, ≂ a = 2 ≂ B, ≂ CED = 2 ≌ B, ≌ ad
2. The line L passing through the focus of 3x + 4Y = 12 intersects the ellipse at two points a and B. if the sum of the distances from two points a and B to the right guide line is 7, the ellipse is straight
Elliptic equation: 3x & sup2; + 4Y & sup2; = 12
x²/4+y²/3=1
a²=4,a=2
b²=3,c²=a²-b²=4-3=1
C=1
e=c/a=1/2
Right guide line: x = A & sup2 / C = 4
Right focus (1,0)
Let a (x1, Y1) B (X2, Y2)
Let y = K (x-1) be substituted into elliptic equation
3x²+4k²(x-1)²=12
(4k²+3)x²-8k²x+4k²-12=0
x1+x2=8k²/(4k²+3)
The sum of distances from points a and B to the right guide line = 4-x1 + 4-x2 = 8 - (x1 + x2)
According to the meaning of the title
8-(x1+x2)=7
8k²/(4k²+3)=1
k²=3/4
k=±√3/2
Y = ± √ 3 / 2 (x-1) means y = √ 3 / 2x - √ 3 / 2 or y = - √ 3 / 2x + √ 3 / 2
Note: if the line passes through the left focus, it is not suitable because the distance from the left focus to the guide line is 5, and the distance from a and B to the guide line is greater than 7
As shown in the figure, in the triangle ABC, the angle ACB = 90 ° is known, and the angle bisector BC of the angle CAD intersects
1. If ∠ B = 40 °, calculate the degree of ∠ AEB. 2. If ∠ B is equal to alpha, use the algebraic formula with alpha to express the degree of ∠ AEB
∠AEB=180º-∠B-∠CAB/2=180º-∠B-(90º-∠B)/2=135º-∠B/2
If ∠ B = 45 degree
Then ∠ AEB = 135 ° - 22.5 ° = 112.5 °
If ∠ B = α, ∠ AEB = 135 ° - α / 2
It is known that a chord ab of ellipse X & # 178 / 9 + Y & # 178 / 4 = 1 is solved by the equation of line AB by point P (1,1)
The solution is derived from the midpoint chord formula of ellipse
KopKab=-b^2/a^2
Let Kop = 1,
be
1 times KAB = - 4 / 9
That is KAB = - 4 / 9
So the equation of line AB is Y-1 = - 4 / 9 (x-1)
That is 9y-9 = - 4x + 4
That is 4x + 9y-13 = 0
It is known that: as shown in the figure, in △ ABC, the bisector BD of ∠ ABC intersects the bisector ce of the outer angle of ∠ ACB at point P
As shown in the figure, in the triangle ABC, the bisector BD of ∠ ABC and the bisector CD of the outer angle of ∠ ACB intersect with D, and it is proved that ∠ d = 1 / 2 ∠ a
It is proved that: ACD = (π - ACB) / 2,
∠DCB=∠ACB+∠ACD=(π+∠ACB)/2
∠D=π-∠DCB-∠CBD=π-(π+∠ACB)/2-∠ABC/2=(π-∠ABC-∠ACB)/2
=∠A/2