It is known that the focus of the ellipse x ^ 2 / 4 + y ^ 2 = 1 is F1, F2, and the intersection of the parabola y ^ 2 = PX (P > 0) and the ellipse in the first quadrant is Q, if ∠ f1qf2 = 60 °, (1) Calculate the area of △ f1qf2; (2) Find the equation of this parabola

It is known that the focus of the ellipse x ^ 2 / 4 + y ^ 2 = 1 is F1, F2, and the intersection of the parabola y ^ 2 = PX (P > 0) and the ellipse in the first quadrant is Q, if ∠ f1qf2 = 60 °, (1) Calculate the area of △ f1qf2; (2) Find the equation of this parabola

Let F1q = m f2q = n. according to the cosine theorem, | F1F2 | ^ 2 = m ^ 2 + n ^ 2-2mncos ∠ f1qf24c ^ 2 = m ^ 2 + n ^ 2 + 2mn-2mn (COS ∠ f1qf2 + 1) 4A ^ 2-4b ^ 2 = (M + n) ^ 2-2mn (COS ∠ f1qf2 + 1), we can get Mn = 2B ^ 2 / (COS ∠ f1qf2 + 1) s △ f1f2q = Mn * sin ∠ f1qf2 / 2 = B ^ 2 * s

The equation for a circle passing through the intersection of ellipse x ^ 2 / 9 + y ^ 2 / 4 = 1 and ellipse x ^ 2 / 4 + y ^ 2 / 9 = 1 is

x^2+y^2=72/13

It is known that hyperbola and ellipse X & # 178; + 4Y = 64 have the same focus, and its asymptote equation is x + √ 3Y = 0

Dear landlord
1) An asymptote equation of ellipse x ^ 2 + 4Y ^ 2 = 64 is x + √ 3Y = 0,
Let the hyperbolic equation be x ^ 2-3y ^ 2 = λ (λ≠ 0),
From C ^ 2 = λ + (λ / 3) = (4 √ 3) ^ 2, λ = 36,
The hyperbolic equation is x ^ 2-3y ^ 2 = 36
I wish you every success

Find the standard equation of hyperbola with the ellipse 3x square + 4Y square = 48 and the real axis length of 2. Hope there is a process, thank you!

The standard equation of ellipse is x ^ 2 / 16 + y ^ 2 / 12 = 1, that is, the focal point is (± 2,0). From the real axis length of hyperbola is 2, we can know that 2A = 2, that is, a = 1 and C = 2. In hyperbola, C ^ 2 = a ^ 2 + B ^ 2 leads to B ^ 2 = 3, so the standard equation of hyperbola is x ^ 2-y ^ 2 / 3
＝1.

Let the ellipse and hyperbola 3x-4y square = 48 have the same focal point, and the major axis is 16, then the standard equation of ellipse is obtained

The focus of hyperbola 3x & # 178; - 4Y & # 178; = 48 is (- 2 √ 7,0), (2 √ 7,0) let the standard equation of ellipse be X & # 178; / A & # 178; + Y & # 178; / B & # 178; = 1 (a > b > 0), then a & # 178; = B & # 178; + 282A = 16  B = 6, the standard equation of ellipse is X & # 178; / 64 + Y & # 178; / 36 = 1, absolutely original! Please add points

Hyperbola and ellipse have the same focus F1 (0, - 5), F2 (0,5), and point P (3,4) is an intersection of asymptote of hyperbola and ellipse

From the common focus F1 (0, - 5), F2 (0,5), let the elliptic equation be y2a2 + x2a2 − 25 = 1, the hyperbolic equation be y2b2 − X225 − B2 = 1, the point P (3,4) is on the ellipse, 16a2 + 9A2 − 25 = 1, A2 = 40, the asymptote of hyperbola passing through point P (3,4) is y = 43x, the analysis has b225 − B2 = 1

It is known that the left and right focus of hyperbola x ^ 2 / A ^ 2-y ^ 2 / b ^ 2 = 1 (a > 0, b > 0) are F1 (- C, 0) and F2 (C, 0), respectively. If there is a point P (different from the endpoint of real axis) on the hyperbola, such that the csin angle pf1f2 = asin angle pf2f1, what is the value range of hyperbolic eccentricity? It needs to be solved

Let p be on the right branch of the hyperbola
According to the sine theorem, because the csin angle pf1f2 = the asin angle pf2f1
c|PF2|=a|PF1|
So C / a = | Pf1 | / | PF2 | = (2a + | PF2 |) / | PF2 | = (2a) / | PF2 | + 1
Another point P (different from the endpoint of the real axis)
So | PF2 | > c-a
So C / A

Given that point P is the intersection of ellipse X21 + A2 + y2a2 = 1 and hyperbola X21 − A2 − y2a2 = 1, F1 and F2 are the focus of ellipse, then cos ∠ f1pf2=______ ．

According to the definition of ellipse, Pf1 + PF2 = 21 + A2. According to the definition of hyperbola, | Pf1 − PF2 | = 21 − A2, 2 (Pf12 + PF22) = 8 can be obtained by adding the squares of the two sides of the above formula at the same time, that is, Pf12 + PF22 = 4 ∵ f2f12 = 4 ∵ Pf12 + pf22f2f12 ∵ cos}

The definition of hyperbola is that the absolute value of the distance difference between any point on the track and two focal points is less than the focal length, so if it is equal to the focal length, what about the image?

When the distance between two fixed points is equal to the focal length, the trajectory is two rays
Two rays end at two points and are on a straight line passing through them

If the right focus of hyperbola x ^ 2 / 9-y ^ 2 / 16 = 1 is F1, point a (9,2), and point m is on the hyperbola, then the minimum value of Ma + 3 / 5mf1

The distance between M and F1 is greater than the distance between M and the right guide line d = e = C / a = 5 / 3
d=3/5MF1
Ma + 3 / 5mf1 = ma + d > = m distance to right guide line = 9-9 / 5 = 36 / 5