What is a parallel string? The original problem: known ellipse x squared / 2 + y squared = 1

What is a parallel string? The original problem: known ellipse x squared / 2 + y squared = 1

Elliptic equation: X & sup2 / 2 + Y & sup2; = 1
Let the intersection of chord and ellipse a (x1, Y1) B (X2, Y2)
Substituting into the elliptic equation x & sup2; + 2Y & sup2; = 2
x1²+2y1²=2
x2²+2y2²=2
Subtraction of two formulas
x1²-x2²+2（y1²-y2²）=0
(x1+x2)(x1-x2)+2(y1-y2)(y1+y2)=0
Let the midpoint be m (x, y)
Then X1 + x2 = 2x, Y1 + y2 = 2Y
According to the meaning (y1-y2) / (x1-x2) = 2
therefore
2x+2×2y×2=0
x+4y=0
It is the locus of the middle point

A triangle and a parallelogram have the same area and height. It is known that the bottom of a triangle is 3 decimeters, and how many decimeters is the bottom of a parallelogram

It is known that the base of triangle is 3 decimeters, and that of parallelogram is 3 × 1 / 2 = 1.5 decimeters

The product of two primes greater than 2 must be (). A. prime B. composite C. even D. odd

The product of two prime numbers greater than 2 must be (B. composite, D. odd)

The circumference of an isosceles triangle is 50 cm, and its waist is twice as long as its bottom. How many cm is the bottom of this triangle?
No equations

Let the length of the bottom edge be x and the waist length be 2x
∴x+2x+2x=50
5x=50
x=10
Waist length: 10 × 2 = 20 (CM)
The bottom length is 10 cm and the waist length is 20 cm
This is not an equation

If AC = 8 and the distance from the center of the circle to the chord is 3, then PA=

Connect OA
Because PA is tangent to circle O, OA is perpendicular to pa
Because ob is perpendicular to AC, △ oba is similar to △ OAP
So, ob / OA = AB / PA
Because the distance from the center of the circle to the chord is 3, that is ob = 3, ab = 4
So in the right angle △ OAB, OA = 5 (Pythagorean theorem)
So 3 / 5 = 4 / PA
So PA = 20 / 3

As shown in the picture, from a to B, there is a small river (both sides are parallel). Now we need to build a bridge on the river (the bridge is perpendicular to the river bank). How to choose the location of the bridge?
As shown in the picture, there is a small river from a to B (both banks are parallel). Now we need to build a bridge on the river (the bridge is perpendicular to the river bank). How to choose the location of the bridge to make the distance from a to B shortest?

This method is correct. As shown in the figure below: the route from a to B is a-m-n-b, and Mn is the fixed value, as long as am + BN is the shortest. Mn = AC, BC is the shortest segment from C to B. Mn is the bridge construction position. ∵ Mn = AC, Mn ∥ AC ∥ quadrilateral amnc is parallelogram ∥ am = NC ∥ am + BN = BN + NC and ∵ B, N, C three points

As shown in the figure, O is a point on the straight line AB, ∠ cod = 90 °, OE bisects ∠ AOC, of bisects ∠ BOD; (1) if ∠ BOC = 40 °, calculate the degree of ∠ EOF; (2) when od bisects ∠ AOF, calculate the degree of ∠ BOC

(1) ∵ - BOC = 40 °, ∵ cod = 90 °, ∵ - AOD = 180 ° - ∵ cod - ∵ BOC = 50 °, ∵ AOC = ∵ AOD + ∵ cod = 140 °, ∵ BOD = ∵ cod + ∵ BOC = 130 °, ∵ OE bisection ∵ AOC, of bisection ∵ BOD, ∵ AOE = 12 ? AOC = 70 °, ? BOF = 12 ∵ BOD = 65 °, ? EOF = 180 ° -

If f (x) = (PX ^ 2 + 2) / (3x + Q) is an odd function and f (2) = 5 / 3, find the value of P and Q
Solution 1: because f (x) = (PX ^ 2 + 2) / (3x + Q) is an odd function
So from F (- x) = - f (x), we get
（px^2+2）/（-3x+q）=-（px^2+2)/(3x+q）
The deformation is (PX ^ 2 + 2) / (- 3x + Q) = (PX ^ 2 + 2) / (- 3x-q)
So q = - Q, q = 0, and f (2) = 5 / 3, P = 2
(is there any mistake in the procedure of finding Q here? Do you want to convert the fractional equation into an integral equation first and then find it)
Solution 2: F (x) = (PX ^ 2 + 2) / (3x + Q) is an odd function, and f (2) = 5 / 3
Then f (2) = (4P + 2) / (6 + Q) = 5 / 3, and then f (2) = (4P + 2) / (6 + Q) = 5 / 3
4P + 2 = 5,6 + q = 3, the solution is p = 3 / 4, q = 3
This solution is obviously wrong, but how to explain it?

F (2) = (4P + 2) / (6 + Q) = 5 / 3, so there is
4p+2=5,6+q=3
This is wrong
Because (4P + 2) / (6 + Q) = 5 / 3 is a quadratic equation
There are countless solutions
If P = 0, q = - 24 / 5
So the solution is wrong

It is known that the line L passing through point P (3,5) intersects with X axis and Y axis with ab respectively. If AP vector = twice Pb vector, the equation of line L is obtained

Let a (a, 0), B (0, b)
Then vector AP = (3-A, 5), vector Pb = (- 3, B-5)
3-A = - 6,5 = 2b-10
A = 9, B = 15 / 2
K(l)=-b/a=-5/6
So the equation of line L is y = - 5x / 6 + 15 / 2
The result is: 5x + 6y-45 = 0

Observe the following monomials: A, - 3A ^ 2,5a ^ 3, - 7a ^ 4,9a ^ 5
(1) According to this rule, write the first monomial in 2012
(2) What is the nth monomial?

-4023a^2012
What is the nth monomial
(-1)^(n-1) * (2n-1)a^n