The left and right focus of the hyperbola x2a2 − y2b2 = 1 is F1, F2, P is a point on the hyperbola, satisfying | PF2 | = | F1F2 |, and the straight line Pf1 is tangent to the circle x2 + y2 = A2, then the eccentricity e of the hyperbola is () A. 3B. 233C. 53D. 54

The left and right focus of the hyperbola x2a2 − y2b2 = 1 is F1, F2, P is a point on the hyperbola, satisfying | PF2 | = | F1F2 |, and the straight line Pf1 is tangent to the circle x2 + y2 = A2, then the eccentricity e of the hyperbola is () A. 3B. 233C. 53D. 54

Let Pf1 be tangent to the circle at point m, and f2h be perpendicular to Pf1 through F2, then h is the midpoint of Pf1, ∵ PF2 | = | F1F2 |, ∵ pf1f2 is isosceles triangle, ∵ f1m | & nbsp; = 14 | & nbsp; Pf1 |, ∵ in right triangle f1mo, | f1m | 2 = c2-a2, | f1m | = b = 14 | Pf1 |, ∵ 2A = 4b-2c ∵ C2 = A2 + B2, ∵ 3C = 5a, ∵ e = 53

Given that the right focus of the hyperbola x2a2-y25 = 1 is (3,0), then the eccentricity of the hyperbola is equal to ()
A. 31414B. 324C. 32D. 43

The right focus of ∵ hyperbola x2a2-y25 = 1 is (3, 0), ∵ A2 + 5 = 9 ∵ A2 = 4 ∵ a = 2 ∵ C = 3 ∵ e = CA = 32, so C is selected

The focus of the ellipse is on the axis, the distance between the two focuses is 12, and the eccentricity is 3 / 4

2c=12 ,c/a=3/4 ,
So C = 6, a = 8,
Then B ^ 2 = a ^ 2-C ^ 2 = 64-36 = 28,
So the equation is x ^ 2 / 64 + y ^ 2 / 28 = 1, or Y ^ 2 / 64 + x ^ 2 / 28 = 1

Given that the eccentricity of hyperbola is 2, the focus is (- 4,0), (4,0), then the hyperbolic equation is______ ．

By E = 2, C = 4, e = Ca, we can get a = 2, B2 = c2-a2, B2 = 12, so the hyperbolic equation is x24-y212 = 1, so the answer is x24-y212 = 1

The focus eccentricity is known and the hyperbolic equation is solved
Focus (- 40) (40) eccentricity 2

Let the hyperbolic equation be x ^ 2 / A ^ 2-y ^ 2 / b ^ 2 = 1
c=4
Eccentricity = C / a = 2 a = 2
a^2+b^2=c^2=16
The simultaneous solution is a ^ 2 = 4, B ^ 2 = 12
The hyperbolic equation is x ^ 2 / 4-y ^ 2 / 12 = 1

The hyperbolic equation with eccentricity of 2 and focus of (0,4) is

c=4,
c/a=2,
A = 2, B ^ 2 = C ^ 2-A ^ 2 = 12, Y type,
The hyperbolic equation is y ^ 2 / 4-x ^ 2 / 12 = 1

It is known that the eccentricity of hyperbola x ^ 2 / A ^ 2-y ^ 2 / b ^ 2 = 1 is √ 3, a Quasilinear equation is x = 1, passing through the right focus F of hyperbola
(1) find the hyperbolic equation (2) find the absolute value of the vector Mn

(1) From C / a = √ 3, a ^ 2 / C = 1, a = √ 3, C = 3, so B = √ 6 hyperbolic equation is x ^ 2 / 3-y ^ 2 / 6 = 1 (2) focus f (3,0), let the line l be x = ty + 3 (t = 1 / k) substituted into hyperbolic equation, 2 (TY + 3) ^ 2-y ^ 2-6 = 0 à (2t ^ 2-1) y ^ 2 + 12ty + 12 = 0, let Y1, Y2 be two of the equation, because FM = - 2fn, so y

How to find the derivative of (LNX) ^ x

Y = (LNX) ^ x = e ^ ln [(LNX) ^ x] = e ^ [XLN (LNX)], then y '= e ^ [XLN (LNX)] * [XLN (LNX)]' = [(LNX) ^ x] * [ln (LNX) + (X / LNX) * (1 / x)] = [(LNX) ^ x] * [ln (LNX) + (1 / LNX)] = [(LNX) ^ x] * [ln (LNX)] + (LNX) ^ (x-1)

The derivative of LNX ^ x =?

Y = x ^ (LNX)
lny=lnx^(lnx)
lny=lnx*lnx
lny=ln²x
(1/y)*y'=2(lnx)*1/x
y'=(2lnx)/x*y
y'=(2lnx)/x*x^lnx
y'=(lnx)*2x^(lnx-1)

Let f (x) = LNX + X, equation 2mf (x) = x ^ 2 have unique real number solution, find the value of positive number M

A:
f(x)=lnx+x,2mf(x)=x^2
2m(lnx+x)=x^2
2mlnx+2mx=x^2
Let g (x) = 2mlnx + 2mx-x ^ 2, x > 0, M > 0
Derivation:
g'(x)=2m/x+2m-2x
The solution G '(x) = 0 is: x-m / x = M
The solution is: X1 = [M + √ (m ^ 2 + 4m)] / 2 (the other x2 does not conform to x > 0, rounding off)
x1-m/x1=m……………………………… （1）
0