Given that P (x, y) is on the ellipse x ^ 2 / 4 + y ^ 2 / 9 = 1, find the maximum value of u = 2x-y

Given that P (x, y) is on the ellipse x ^ 2 / 4 + y ^ 2 / 9 = 1, find the maximum value of u = 2x-y

Using parametric equation
x=2cosp
y=3sinp
Then u = - 3sinp + 4cosp
=-(3sinp-4cosp)
=-√(3²+4²)sin(p-q)
=-5sin(p-q)
Where tanq = 4 / 3
So the maximum value is 5

It is known that the center of the ellipse C is at the coordinate origin, the focus is on the x-axis, the eccentricity is 1 / 2, and the maximum distance between the point on the ellipse C and the focus is 3

According to the meaning of the title, a = 2c, a + C = 3
So a = 2, C = 1
The standard equation is
X^2/4+Y^2/3=1

If the circumscribed circle radius of an isosceles right triangle is 1, then its inscribed circle radius is 1
If the circumscribed circle radius of an isosceles right triangle is 1, then its inscribed circle radius is 1

=(radical 2) - 1
Isosceles right triangle three sides length is 2, root sign 2, root sign 2
The area is 1
The radius of inscribed circle is 2 * area / perimeter = 2 * 1 / (2 + 2 root sign 2) = root sign 2-1

Let u = R, a = {x | 0 ≤ x ＜ 3}, B = {x | - 1 ＜ x ≤ 2}. Find Cu (AUB), CUA, cub

A={x|0≤x＜3},B={x|-1＜x≤2}
Then a ∪ B = {x | - 1

As shown in the figure, in the triangle ABC, AC = BC angle ACB = 90 degree, D is the point on AC, and AE is perpendicular to the extension line of BD, and AE = 1 / 2bd
It is known that in the triangle ABC, AC = BC, ACB = 90 degrees, D is the point on AC, and AE is perpendicular to the extension line of BD, and AE = 1 / 2bd
Verification: BD is the bisector of angle ABC

Prolonging the intersection of AE and BC in F
∠EAD=∠CBD
∠ACF=∠ACB=90
AC=BC
ΔACF≌ΔBCD
∴AF=BD
And AE = 1 / 2bd
∴EF=1/2BD=AE
ΔABE≌ΔFBE
∴∠ABD=∠CBD

If the equation 2aX & # 178; - X-1 = 0 has a solution in the interval (0,1), then the value range of a is

Let f (x) = 2aX ^ 2-x-1
Then f (0) * f (1) 1

Finding indefinite integral ∫ 10 ^ (2arccos x) / √ (1-x ^ 2) DX

-§-(10^(2arccosx)/((1-x^2)^(1/2))dx=-1/2§10^(2arccosx)d(2arccosx)=-(1/2)10^(2arccosx)/In|0+C

The distance between the two places is 1800 meters. Party A and Party B walk from the two places at the same time and meet each other in 12 minutes (speed a is faster than speed b). If each person walks 25 meters more per minute, the distance between the meeting place and the last meeting point is 33 meters. What is the speed of Party A and Party B?

The meeting time of a and B after increasing speed is: 1800 △ 12 + 25 × 2, = 1800 △ 200, = 9 (minutes); suppose a's speed is x meters per minute, according to the title: 12x-9 (x + 25) = 33, & nbsp; 12x-9x-225 = 33, & nbsp; 3x-225 + 225 = 33 + 225 & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; 3x = 258; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; X = 86, then B's speed is 1800 △ 12-86 = 64 (m); answer: A's speed is 86 M / min, B's speed is 64 m / min

How to use the triangle method of force balance similarity
Suitable for those topics, what is the scope of application

It is suitable for judging the direction and size of vector calculation; all vectors are connected end to end in turn; the direction of the last first quantity and the last quantity is the direction of the vector; the size is the size of the vector

For a rectangle, if its length is reduced by 8 meters, its area will be reduced by 480 square meters; if its width is increased by 6 meters, its area will be increased by 240 square meters?

The original area is 2400 square meters
1.480/8 = 60m. This is the original width
240 / 6 = 40 meters. This is the original length
3.40 * 60 = 2400 square meters. The original area