On the number axis, all integers whose distance to the origin is not more than 3 are If 0

-3,-2,-1,0,1,2,3,
-1 of a < - a < a < 1 of a
two
c

How many integer points on the number axis whose distance from the origin is greater than 2 and not greater than 4

Integers greater than 2 and not greater than 4
-4=

When the focus of an ordinary elliptic equation is on the Y. axis, how can it be transformed into a parametric equation?
In addition, what's the difference from the focus on the x-axis

No difference, x ^ 2 / b ^ 2 + y ^ 2 / A ^ 2 = 1, can be written as parameter equation:
x=bcost
y=asint

The focus of the ellipse is on the y-axis. What is the equation of the ellipse

The extremum of one variable equation in MATLAB
X=-9.8933+24.2405T-16.9282T^2+1.7470T^3
X range 0

XM = 0; XM = 6; x0 = 1;%% find the minimum ftouchby = inline ('- 9.8933 + 24.2405 * t-16.9282 * t. ^ 2 + 1.7470 * t. ^ 3','t '); [Tmin, Fmin] = fmincon (ftouchby, x0, [], [], [], [], XM, XM,');% fmincon is find the minimum%%, find the maximum ftouchby_ 1=inline('-1*(-9....

How to find the extremum of a fractional equation?
For example, how to find the extremum of (2-x) x / x-4?

Let y = primitive = (2x - x ^ 2) / (x - 4)
xy - 4y + x^2 -2x = 0;
x^2 + (y-2)x - 4y = 0
delta = (y-2)^2 + 16y >=0
so ,y^2 +12y + 4 >=0
Y is its range

Let the function y = y (x) be determined by the parametric equation {x = 1-3T - (T ^ 3) y = 1-3T + (T ^ 3)}, and find the extremum of the function

dx/dt=-3-3t²
dy/dt=-3+3t²
y'=(dy/dt)/(dx/dt)=(-3+3t²)/(-3-3t²)=(1-t²)/(1+t²)
From y '= 0, t = 1, - 1
dy'/dt=-4t/(1+t²)²
y"=-4t/(1+t²)²/(-3-3t²)=(4t/3)/(1+t²)³
When t = - 1, y“

It is known that the center of the ellipse is at the origin and the focus is on the x-axis
The eccentricity e = 2, its intersection with the straight line x + y + 1 = 0 is p, Q, and the circle with PQ as the diameter passes through the origin to solve the elliptic equation

The eccentricity of ellipse e belongs to (0,1). No wonder you don't. the title is wrong

The center of the ellipse is at the origin, the focus is on the x-axis, the line between a focus and the two ends of the minor axis is perpendicular to each other, and the distance between the focus and the nearer end of the major axis is root 10-root 5, so the elliptic equation can be solved

A focal point is perpendicular to the two ends of the minor axis
Then the angle between the line of a focal point and the end of a minor axis and the major axis is 45 degrees
That is, B = C
a^2=b^2+c^2=2c^2
a=√2c
The distance between the focus and the near end of the long axis is √ 10 - √ 5
So a-c = √ 10 - √ 5
(√2-1)c=√10-√5
So C = √ 5
a=√10
b^2=c^2=5
a^2=10
So x ^ 2 / 10 + y ^ 2 / 5 = 1

It is known that the center of the ellipse e is at the origin, the focus is on the x-axis, e = 1 / 2, and the sum of the distances from one point on e to two focuses is 4. The left focus F1 passing through the ellipse e is a straight line L intersecting with the ellipse and two points ab. if the area of the triangle AOB is equal to 2 / 7 of 6 root signs, the equation of the straight line L is obtained

In an ellipse, 2A = 4, a = 2, C / a = e = 1 / 2, C = 1, B ^ 2 = a ^ 2-C ^ 2 = 3, the e-equation of the ellipse is x ^ 2 / 4 + y ^ 2 / 3 = 1. ① F1 (- 1,0), let the equation of l be x = MY-1, and substitute it into ① * 12,3 (m ^ 2Y ^ 2-2my + 1) + 4Y ^ 2 = 12, then (3m ^ 2 + 4) y ^ 2-6my-9 = 0, △ = 36m ^ 2 + 36 (3m ^ 2 + 4) = 144 (m ^ 2 + 1), | ab | = √

On the number axis, all integers whose distance to the origin is not more than 3 are If 0