The distance from the point P on the left branch of hyperbola X & sup2 / 64-y & sup2 / 36 = 1 to the left quasilinear is 10, so what is the distance from P to its right focus

The distance from the point P on the left branch of hyperbola X & sup2 / 64-y & sup2 / 36 = 1 to the left quasilinear is 10, so what is the distance from P to its right focus

Is the equation of hyperbola: X / 64-y / 36 = 1? If yes, then a = 8, B = 6  C = 10, the two quasars are: x = positive and negative, a / C = positive and negative 6.4 〉 the left quasars are x = - 6.4, the right quasars are x = 6.4, the distance between the quasars is 12.8 ∵ the distance from point P to the left quasars is 10 ∵ and the distance from point P to the right quasars is 12.8-10 = 2.8

If the distance between the point P and L on the hyperbola is 5 / 3, find the trajectory of P

a²=4,b²=5
Then C & # 178; = 9
c=3
So l is x = A & # / C = 4 / 3
So if the distance to L is 5 / 3, then the abscissa is 4 / 3-5 / 3 = - 1 / 3 or 4 / 3 + 5 / 3 = 3
Because a = 2
So, in the point on the hyperbola, | x | ≥ 2
So the abscissa is 3
I'll take y
So p has two, namely (3, ± √ 5 / 2)

Hyperbolic equation x square / 16-y square / 9 = 1, the distance from a point to the right focus is 4, find the distance from P to the right quasilinear and to the left quasilinear
It's best to have a description, 3Q

Thinking; using the second definition to solve
Let the distance to the right guide line be D, then there are 4 / D = 5 / 4, d = 16 / 5. Thus the distance to the left guide line is 48 / 5

What is the inflection point of the curve y = 1 + X / (x + 3) ^ 2?

y'=1/(x+3)^2-2*x/(x+3)^3
y''=-4/(x+3)^3+6*x/(x+3)^4
Let y '' = 0 get x = 6
Inflection point: (6,87 / 81)

Inflection point coordinates (x, y) of curve y = (1 / 3) x ^ 3-x ^ 2 + 1=

y=(1/3)x^3-x^2+1
y'=x^2-2x
y''=2x-2
Inflection point coordinates (x, y) = (1,1 / 3)

Find the inflection point of function curve y = x ^ 3 (1-x),

The inflection point should be consistent with the fact that both the first and second derivatives are 0, while the third derivative is not 0
So y = x ^ 3-x ^ 4
y'=3x^2-4x^3=0,x=0,x=3/4
y"=6x-12x^2=0,x=0,x=1/2
When y (3) = 6-24x and x = 0, y (3) is not equal to 0
So x = 0 is the inflection point
So the inflection point is (0,0)

Finding the extremum of function f (x, y) = XE - (x ^ 2 + y ^ 2) / 2

The extreme value of the function f (x (x, y) is the extreme value of the function f (x (x, y) of the extreme value of the extreme value of the function f (x \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\35; 8706; X # 8706; y = 0; C = # 870

What is the minimum value of the function y = Xe ^ x

y = xe^x
dy/dx = xe^x + e^x = e^x (x + 1)
dy/dx = 0
e^x (x + 1) = 0
X + 1 = 0 or (e ^ x = 0, rounding)
x = -1
The minimum value is y = (- 1) e ^ (- 1) = - 1 / E when x = - 1

The function determined by y-xe ^ y = 2 is y = y (x), and Y is calculated“

Y-xe ^ y = 2Y '- e ^ y-xe ^ y * y' = 0y '(1-xe ^ y) = e ^ y 1y' = e ^ y / (1-xe ^ y) by deriving Formula 1, y '' (1-xe ^ y) + y '(- e ^ y-xe ^ y * y') = e ^ y * y '(1-xe ^ y) = e ^ y * y' + e ^ YY '+ Xe ^ y (y') ^ 2 = 2E ^ y * y '+ Xe ^ y (y') ^ 2 = e ^ y (2Y '+ (y') ^ 2) = e ^ y (2e ^ y / (1-xe ^ y) + (e ^ y / (...)

The inflection point of the curve y = 3x ^ 4-4x ^ 3 is
To use the problem-solving process, and talk about the inflection point method. Thank you

Derivation
y'=12x^3-12x^2
y''=36x^2-24x
The inflection point is y '' = 0
So x = 0, x = 2 / 3
y''=12x(3x-2)
Then x = 0 and x = 2 / 3
There are y "signs on both sides of the point
So it's all inflection points
So it's (2 / 3, - 16 / 27), (0,0)