It is known that ellipse x2 / A2 + Y2 / B2 = 1 (a > b > 0) and hyperbola x2 / m2-y2 / N2 = 1 (m, n > 0) have common focus F1, F2, P are an intersection point. The proof is: angle f1pf2 = 2arctan (n / b) triangle f1pf2 area = BN

It is known that ellipse x2 / A2 + Y2 / B2 = 1 (a > b > 0) and hyperbola x2 / m2-y2 / N2 = 1 (m, n > 0) have common focus F1, F2, P are an intersection point. The proof is: angle f1pf2 = 2arctan (n / b) triangle f1pf2 area = BN

Let p be at the right branch of the hyperbola
|PF1|+|PF2|=2a①,|PF1|-|PF2|=2m②,
From ① and ②, we can get: | Pf1 | = a + m, | PF2 | = a-m,
And a & # 178; - B & # 178; = M & # 178; + n & # 178; = C & # 178;,
∴cos∠F1PF2=(|PF1|²+|PF2|²-|F1F2|²)/2|PF1|·lPF2l
=[(a+m)²+(a-m)²-(2c)²]/2(a+m)(a-m)
=[(a²-c²)+(m²-c²)]/(a²-m²)
=(b²-n²)/(b²+n²),sin∠F1PF2=2bn/(b²+n²),
∴tan½∠F1PF2=(1-cos∠F1PF2)/sin∠F1PF=n/b,
∴∠F1PF2=2arc tan(n/b),
Δ f1pf2 area = (?) 189; lpf1l · lpf2l · sin ∠ f1pf2
=½(a+m)(a-m)·2bn/(b²+n²)
=bn(a²-m²)/(a²-m²)=bn.
It is known that ellipse C1: x2 / A2 + Y2 / B2 = 1 (a > b > 0) and hyperbola C2: x2
It is known that the ellipse C1: x2 / A2 + Y2 / B2 = 1 (A & gt; B & gt; 0) and the hyperbola C2: x2-y2 / 4 = 1 have a common focus, and an asymptote of C2 intersects with the circle with the diameter of the major axis of C1 at two points A.B. if C1 exactly divides the line AB into three equal parts, B ^ 2 = 0.5, the focus of C2 is (± √ 5,0), an asymptote equation is y = 2x, according to the symmetry, AB is the diameter of the circle, and ab = 2A, the half focus of C1 is C = √ 5, So we get a ^ 2-B ^ 2 = 5. ① let the coordinates of the intersection of C1 and y = 2x in the first quadrant be (x, 2x), and substitute it into the equation of C1 to get: x ^ 2 = (a ^ 2B ^ 2) / (b ^ 2 + 4A ^ 2) ②. From the symmetry, we know that the chord length of the straight line y = 2x cut by C1 = 2x √ 5. From the question, we get: 2x √ 5 = 2A / 3, So x = A / (3 √ 5) ③ a ^ 2 = 11b ^ 2 is obtained from ② ③ ④ a ^ 2 = 5.5 B ^ 2 = 0.5 is obtained from ① ④, where [from the symmetry, the chord length of the straight line y = 2x cut by C1 = 2x √ 5] how to understand... Very depressed, how to multiply 2 by the root sign 5 by X
Let the coordinates of the intersection of C1 and y = 2x in the first quadrant be d (x, 2x)
So od = root (x ^ 2 + 4x ^ 2) = root 5 * X
Therefore, the symmetry shows that the chord length cut by the intersection of the line y = 2x and C1 = 2od = 2x * radical 5
It is known that a Quasilinear equation of ellipse C1: x2 / A2 + Y2 / B2 is x = 25 / 4, and its left and right vertices are a and B respectively. Hyperbola C2: x2 / a2-y2 / B2 = 1, and an asymptote equation of hyperbola is 3x-5y = 0
Q: take a point P on hyperbola C2 in the first quadrant, connect AP intersection ellipse C1 to point m, connect Pb and extend intersection ellipse C1 to point n. if am vector = MP vector, prove: Mn vector * AB vector = 0
From the known a 2C = 25 4 B a = 35 C 2 = a 2-B 2 solution, it is obtained that the equation of a = 5 B = 3 C = 4 ﹣ ellipse is x 225 + y 29 = 1, the equation of hyperbola is x 225 - y 29 = 1, and C ′ = 25 + 9 = 34 ﹣ the eccentricity of hyperbola is E 2 = 34 5
If the hyperbola x2 / a-y2 / b = 1 (a > 0, b > 0) and the ellipse x2 / M + Y2 / N = 1 (M > n > 0) have the same focus F1, F2. P is an intersection of the two curves
If there are two hyperbolic curves | 1 / 2 | (PFA / M 2) and the common focus of the two hyperbolic curves |, then how many are they
P is on the ellipse
So Pf1 + PF2 = 2 √ a
P on hyperbola
|PF1-PF2|=2√m
PF1-PF2=±2√m
If pf1-pf2 = 2 √ M
PF1+PF2=2√a
PF1=√m+√a
PF2=√a-√m
PF1×PF2=a-m
If pf1-pf2 = - 2 √ M
PF1+PF2=2√a
PF1=√a+√m
PF2=√a+√m
PF1×PF2=a-m
To sum up
PF1×PF2=a-m
If the ellipse = 1 (a > b > 0) and hyperbola = 1 (M > 0, n > 0) have the same focal points, F 1, F 2, P are an intersection of the two curves, then|
If the ellipse x ^ 2 / M + y ^ 2 / N = 1 (M > n > 0) and hyperbola x ^ 2 / S-Y ^ 2 / T = 1 (s > 0, t > 0) have the same focus, F1, F2, P are an intersection of two curves, then | Pf1 | · | PF2|=
Additional bonus points will be given to those who answer questions within one hour
If the ellipse x ^ 2 / M + y ^ 2 / N = 1 (M > n > 0) and hyperbola x ^ 2 / S-Y ^ 2 / T = 1 (s > 0, t > 0) have the same focus, F1, F2, P is an intersection of two curves,
|Definition of Pf1 | + | PF2 | = 2sqrt (m) ellipse
||Definition of hyperbola Pf1 | - | PF2 | = 2sqrt (s)
|PF1|·|PF2|=[（|PF1|+|PF2|）^2- (|PF1|-|PF2 |)^2]/4 = m-s
PF1+PF2=2a PF1-PF2 =2m
So Pf1 = m + a PF2 = A-M
So Pf1 * PF2 = (a + m) (A-M)
If the ellipse x ^ 2 / M + y ^ 2 / N = 1 (M > n > 0) and hyperbola x ^ 2 / S-Y ^ 2 / T = 1 (s > 0, t > 0) have the same focus, F1, F2, P are an intersection of two curves, then | Pf1 | · | PF2|=
Additional bonus points will be given to those who answer questions within one hour
|PF1|+|PF2| = 2√(m)
|PF1|-|PF2| = 2√(s)
|PF1|·|PF2|=
((|PF1|+|PF2|)^2-(|PF1|+|PF2|)^2)/4= m-s
According to the first definition of ellipse:
Pf1 + PF2 = 2A, where a = SQR (m) (SQR is the root sign),
Similarly, hyperbola is defined as:
Pf1-pf2 = 2A, where a = SQR (s)
By subtracting the square of the above two formulas, we can get the following result:
4｜PF1｜* ｜PF2｜=4m-4s
Namely
｜PF1｜* ｜PF2｜=m-s
Dizzy. I don't have enough time to post it!
From the definition of ellipse and hyperbola
|Pf1 | + | PF2 | = 2m ^ 0.5 -- Formula 1
||Pf1 | - | PF2 | = 2S ^ 0.5 -- Formula 2
Square of Formula 1 minus square of formula 2
4|PF1|·|PF2|=4（m-s)
So | Pf1 · | PF2 | = (M-S)
√27-15√1／3+1／4√48
√27-15√(1/3)+1/4*√48
=√(3^3*3)-15√(3/3^2)+1/4*√4^2*3)
=3√3-(15/3)√3+(4/4)√3
=3√3-5√3+√3
=-√3
If M and N are the two real roots of the equation x ^ 2-2ax + 9 about X, then the minimum value of (m-1) ^ 2 + (n-1) ^ 2 is (process)
From the relation between root and coefficient, M + n = 2A, Mn = 9
（m-1)^2+(n-1)^2
=m^2-2m+1+n^2-2n+1
=(m^2+n^2)-2(m+n)+2
=(m+n)^2-2mn-2(m+n)+2
If M + n = 2A, Mn = 9 generations, we can get that,
Original formula = 4A ^ 2-18-4a + 2
=4a^2-4a-16
On the equation AX & # 178; + 2x + 2A = 0 (1) if one of the equations is larger than 1 and the other is smaller than 1, the range of a is obtained
② If the roots of the equation are all greater than 1, the range of a is obtained
First of all, the discriminant = 4-8a ^ 2 > = 0, so - √ 2 / 20, ---- ③
The range of a is - √ 2 / 2
Application of square root
The relationship between the height h (unit: m) of a freely falling object and the falling time t (unit: s) is h = 4.9t & sup2; as shown in the figure, how long does it take for an object to fall freely from a 120m high building to reach the ground (the result is taken as an integer)?
I haven't learned the double root sign
Since H = 4.9t & sup2;, then H = 120m, so t & sup2; = 120 / 4.9
So t = 20 times root sign 3 / 7
I hope I can help you
120=4.9t^2
t^2=120/4.9=1200/49=400*3/49
t=20√3/7
Does the equation x & # 178; + 2aX + A & # 178; = 4 of X have two real roots? Why?
x²+2ax+a²=4
△=4a²-4x1x(a²-4)
=16>0
There are two unequal real roots