F 1 and F 2 are the left and right focus of the hyperbola x ^ 2-y ^ 2 / 3 = 1, m (6,6) is a point inside the hyperbola, P is a point on the right branch of the hyperbola Find the minimum value of 〡 Pf1 〡 + 〡 PF2 〡

F 1 and F 2 are the left and right focus of the hyperbola x ^ 2-y ^ 2 / 3 = 1, m (6,6) is a point inside the hyperbola, P is a point on the right branch of the hyperbola Find the minimum value of 〡 Pf1 〡 + 〡 PF2 〡

If the title is correct, then the minimum value of 〡 Pf1 〡 + 〡 PF2 〡 is equal to the distance between the two focal points. At this time, point P is the intersection of the two focal lines and the right branch. From the meaning of the title, we know that the real half axis is 1, and the imaginary half axis is √ 3, so the half focal length is 2, so the focal length is 4

The tangent equation of parameter equation x = T ^ 3-8t y = T ^ 2 (t is parameter) at point (7,1) is
The key is how to transform the parametric equation into a general equation
If there are other magic methods, you can also show them

Find out the slope of the tangent equation: divide the derivative of y to t by the derivative of X to t, that is 2T / (3T ^ 2-8) (*)
Substituting (7,1) into the equation, t = - 1
Substituting (*) gives a slope of 2 / 5
Then the tangent equation is y = 2 / 5x + C
Then tangent through (7,1) to get: C = - 9 / 5
The tangent equation is y = 2 / 5x-9 / 5
That is: 2x-5y-9 = 0

Ordinary equation of parameter equation x = et + e − ty = 2 (ET − e − T) (t is parameter)______ ．

From the parametric equation, we can get 2x = 2et + 2e − T & nbsp; & nbsp; & nbsp; & nbsp; ① y = 2et − 2e − T & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; ②, by subtracting the squares of ① and ②, we can get 4x2-y216, that is, x24 − y216 = 1, so the answer is: x24 − y216 = 1

The graph of equation x = et + e − ty = et − e − t (t is a parameter) is ()
A. Left branch of hyperbola B. right branch of hyperbola C. upper branch of hyperbola D. lower branch of hyperbola

From the square of x = et + E-T, we can get x2 = e2t + e-2t + 2, and substitute y = et-e-t to get y2 = e2t + e-2t-2. By subtracting the two formulas, we can get x2-y2 = 4, and x = x = et + E-T ≥ 2et · e − t = 2, so the common equation is: x2-y2 = 4 (x ≥ 2), and the graph is the right branch of hyperbola. So we choose B