There is no intersection between the circle whose diameter is the chord of the conic curve passing through the focus and the corresponding guide line. What is the conic curve 1. Ellipse 2. Hyperbola 3. Parabola 4. Not sure

There is no intersection between the circle whose diameter is the chord of the conic curve passing through the focus and the corresponding guide line. What is the conic curve 1. Ellipse 2. Hyperbola 3. Parabola 4. Not sure

The radius of the circle whose diameter is the chord of the ellipse passing through the focus is "the distance to the fixed point (focus)". The distance between the center of the circle and the corresponding guide line is "the distance to the fixed line (corresponding guide line)". There is no intersection point between the circle and the straight line, that is: "the distance to the fixed point (focus)" is less than "the distance to the fixed line (corresponding guide line)"

What is the eccentricity formula of hyperbola

e＝c/a,c²＝a²＋b²

Can you tell me the formula of eccentricity of ellipse?

e=c/a
F is the distance from a point on the ellipse to the focus, D is the distance from the point to the guide line on the other side of the same focus
be
e=f/d

Eccentricity formula of ellipse e = C / a conversion = radical (1 - (B / a) ^ 2)

a²=b²+c²
c²=a²-b²
c=√(a²-b²)
e=c/a=√[(a²-b²)/a²]=√[1-(b/a)²]
So transform

If there is any point P on the hyperbola x2 / a2-y2 / B2 = 1, F1F2 is the focal point of the hyperbola, and the angle f1pf2 =, what is the area of f1pf2

For convenience, let Pf1 = m, PF2 = n, then | M-N | = 2A, and in the triangle pf1f2, there is (2C) & sup2; = M & sup2; + n & sup2; - 2mncos θ = (m-n) & sup2; + 2mn-2mncos θ = 4A & sup2; + 1-cos θ) × 2Mn, that is, (1-cos θ) × 2Mn = 4B & sup2;, and the area of the triangle pf1f2 is s = (1 / 2)

If F1F2 is a hyperbola, the two focal points P of x2-4y2 = - 4 are on the hyperbola and ∠ f1pf2 = 90 °, then the triangle f1pf2 is a surface

The hyperbolic standard equation is: y ^ 2-x ^ 2 / 4 = 1. A = 1, C = √ 5, the focus is on the Y axis. F1F2 = 2C = 2 √ 5, pf1-pf2 = 2A = 2, Pf1 ^ 2 + PF2 ^ 2 = F1F2 ^ 2 = 20. (pf1-pf2) ^ 2 = Pf1 ^ 2 + PF2 ^ 2-2pf1 * PF2 = 20-2pf1 * PF2 = 4A ^ 2 = 4. Then Pf1 * PF2 = 8. The area of triangle f1pf2 = (1 / 2) Pf1 * PF2 = 4

Let F1F2 be the two focal points of hyperbola x ^ 2 / 4-y ^ 2 = 1, point p be on hyperbola and satisfy vector PF2 * vector PF2 = 0, then the area of triangle f1pf2 is
Given that hyperbola C: y ^ 2 / 9-x ^ 2 / 8 = 1, parabola has the lower vertex of curve C as the focus and the origin as the vertex, the standard equation of parabola is obtained

1. A & sup2; = 4A = 2 let Pf1 = m, PF2 = n, then | M-N | = 2A = 4 square M & sup2; + n & sup2; - 2Mn = 16b & sup2; = 1C & sup2; = 4 + 1 = 5, so F1F2 = 2C = 2 √ 5, because vector Pf1 * vector PF2 = 0, so Pf1 is vertical to PF2, so this is a right triangle, area = Mn / 2, Pythagorean theorem M & sup2; + n & sup2; = (2C) & sup2

Let F1F2 be the two focuses of the hyperbola x ^ 2 / A ^ 2-y ^ 2 / b ^ 2 = 1, p be on the hyperbola, if vector Pf1 * vector PF2 = 0, absolute value Pf1 * absolute value PF2 = 2Ac
C is the half focal length, and the eccentricity is calculated

According to the definition of hyperbola, | pf1-pf2 | = 2A, F1F2 = 2c, because the vector Pf1 * vector PF2 = 0, Pf1 * PF2 = 2Ac, so Pf1 ^ 2 + PF2 ^ 2 = F1F2 ^ 2 = (2C) ^ 2 | pf1-pf2 | ^ 2 = (2a) ^ 2, get C ^ 2-ac-a ^ 2 = 0, at the same time, get e ^ 2-e-1 = 0 (E > 1) by a ^ 2, finally e = (1 + radical 5) / 2

On the hyperbola x ^ 2 / A ^ 2-y ^ 2 / b ^ 2 = 1 with E = 5 / 4, F1F2 is its focus, and the vector Pf1 * PF2 = 0. If the area of the triangle f1pf2 is 9, find the value of a + B
Thank you

Plus or minus 2.625

Let F1F2 be the two focuses of the hyperbola x ^ 2 / 4-y ^ 2 = 1, and p be on the hyperbola. When the area of △ f1pf2 is 1, what is the vector Pf1 and the vector PF2 equal to

Let {an} be an arithmetic sequence, {BN} be an arithmetic sequence with all positive numbers and A1 = B1 = 1, A3 + B5 = 21, A5 + B3 = 13