The chord length of the line y = X-1 / 2 cut by the ellipse x ^ 2 + 4Y ^ 2 = 4 is

By eliminating y from {y = X-1 / 2 {X & # 178; + 4Y & # 178; = 4, we get: X & # 178; + 4 (x-1 / 2) & # 178; = 4, we get: 5x & # 178; - 4x-3 = 0. Obviously, Δ = (- 4) & # 178; - 4 × 5 × (- 3) = 76 ＞ 0. By Weida's theorem, we get: X1 + x2 = 4 / 5x1 & # 8226; x2 = - 3 / 5, let y = X-1 / 2 be

The chord length of ellipse x2 + 4y2 = 16 cut by straight line y = x + 1 is______ ．

Substituting the straight line y = x + 1 into the equation of ellipse x2 + 4y2 = 16, we get 5x2 + 8x-12 = 0. Let the intersection of the straight line and the ellipse be a (x1, Y1), B (X2, Y2).. X1 + x2 = - 1.6, x1x2 = - 2.4  the chord length of the ellipse cut by the straight line be ab = 1 + 1 ×| x1-x2 | = 2 × 1.62 + 4 × 2.4 = 43825. So the answer is: 43825

The chord length of ellipse x2 + 4y2 = 16 cut by straight line y = x + 1 is______ ．

Square / 4 of ellipse x plus y square / 3 = 1, the line y is equal to x plus 1 and intersects ellipse at a, B, find the chord length of ab

Let a (x1, Y1), B (X2, Y2) | ab | = √ [(x1-x2) ^ 2 + (y1-y2) ^ 2]. The distance formula ? y1-y2 = K (x1-x2) = x1-x2 | ab | = √ [(x1-x2) ^ 2 + (y1-y2) ^ 2]. Let a (x1, Y1), B (X2, Y2) | ab | = √ [(x1 + x2) ^ 2 + (y1-y2) ^ 2]

Let's know the ellipse G: X6 ^ 2 / 4 + y ^ 2 = 1. Make the tangent l of circle x ^ 2 + y ^ 2 = 1 through point (m, 0) intersect the ellipse g at two points a and B. (1) find the focus coordinates and distance of ellipse G

a^2=4, b^2=1, c^2=3
Focus coordinates (radical 3,0) and (- radical 3,0)
Eccentricity C / a = root 3 / 2

Given that the ellipse x ^ 2 / A ^ 2 + y ^ 2 / (a ^ 2-1) = 1 and the straight line y = X-1 intersect at two points AB, and the circle with ab as the diameter passes through the left focus of the ellipse, the value of a ^ 2 is calculated

The intersection of ellipse and line can be solved by two equations
x²/a²+y²/(a²-1)=1.(1)
y=x-1.(2)
(2) Substitute (1) to get
x²/a²+(x-1)²/(a²-1)=1
Well organized
(2a²-1)x²-2a²x+(2-a²)a²=0
We know that X1 + x2 = 2A & sup2; / (2a & sup2; - 1) from Weida's theorem
x1*x2=(2-a²)a²/(2a²-1)
In the same way, from (2)
x=y+1.(3)
Substitute (1) to get
(y+1)²/a²+y²/(a²-1)=1
(2a²-1)y²+2(a²-1)y-(a²-1)²=0
y1+y2=-2(a²-1)/(2a²-1)
y1*y2=-(a²-1)²/(2a²-1)
Ellipse focal length C = √ [A & sup2; - (A & sup2; - 1)] = 1
The left focus C of the ellipse is (- 1,0)
AC⊥BC
Vector AC = (x1 + 1, Y1)
Vector BC = (x2 + 1, Y2)
Vector ac * vector BC = 0
That is, (x1 + 1) (x2 + 1) + y1y2 = 0
x1*x2+(x1+x2)+1+y1*y2=0
∴(2-a²)a²/(2a²-1)+2a²/(2a²-1)+1-(a²-1)²/(2a²-1)=0
That is, (2-A & sup2;) a & sup2; + 2A & sup2; + 2A & sup2; - 1 - (A & sup2; - 1) & sup2; = 0
Well organized
-2(a²)²+8a²-2=0
(a²)²-4a²+1=0
a²=[4+√(16-4)]/2=2+√3
or
a²=[4-√(16-4)]/2=2-√3

It is known that the line y = X-1 and the ellipse x2m + y2m − 1 = 1 intersect at two points a and B. If a circle with diameter AB passes through the left focus of the ellipse, the value of M is obtained

From the meaning, A2 = m, B2 = M-1, C2 = 1, simultaneous linear equation and elliptic equation, we can get x2m + y2m − 1 = 1y = x − 1, eliminate y and simplify, (2m-1) x2-2mx + 2m-m2 = 0, let a (x1, Y1), B (X2, Y2), from Weida's theorem, X1 + x2 = 2m2m − 1, x1x2 = 2m − m22m − 1; ∵ AF1 · BF1

Given that the focus of the ellipse x2a2 + y2b2 = 1 (a > b > 0) is the same as that of the hyperbola x24-y212 = 1, and the sum of the distances from any point on the ellipse to the two focuses is 10, then the eccentricity of the ellipse is equal to ()
A. 35B. 45C. 54D. 34

∵ the focus coordinates F1 (- 4,0), F2 (4,0) of hyperbola x24 − y212 = 1, ∵ the focus coordinates F1 (- 4,0), F2 (4,0) of ellipse, ∵ the sum of distances from any point on ellipse to two focuses is 10, ∵ 2A = 10, a = 5, ∵ the eccentricity of ellipse e = CA = 45

It is known that ellipse X & # 178 / / M + Y & # 178; = 1 (M > 1) and hyperbola X & # 178 / / n-y & # 178; = 1 (n > 1) have the same focus f1.f2, and P is their intersection point. Then what is the shape of triangle f1pf2?

All afternoon,

Let the two focuses of the ellipse x2132 + y2122 = 1 be F1 and F2. If the absolute value of the distance difference between the moving point on the hyperbola C and F1 and F2 is 8, then the hyperbolic equation is ()
A. x2132−y252＝1B. x2132−y2122＝1C. x232−y242＝1D. x242−y232＝1

According to the definition of hyperbola and | Pf1 | - | PF2 | = 8, we can know that 2A = 8, a = 4, B = 3. The equation of hyperbola is & nbsp; x242 − y232 = 1

The chord length of the line y = X-1 / 2 cut by the ellipse x ^ 2 + 4Y ^ 2 = 4 is