Given that the ellipse x ^ 2 / 25 + y ^ 2 / 9 = 1, the straight line passing through the right focus of the ellipse intersects the ellipse at two points a and B, and intersects the Y axis at point P, let PA vector = k1af vector, Pb vector = k2bf vector, then what is the value of K1 + K2?

Given that the ellipse x ^ 2 / 25 + y ^ 2 / 9 = 1, the straight line passing through the right focus of the ellipse intersects the ellipse at two points a and B, and intersects the Y axis at point P, let PA vector = k1af vector, Pb vector = k2bf vector, then what is the value of K1 + K2?

Let y = KX + (- 4K) = kx-4k, let a (x1, Y1) B (X2, Y2) and P (0, - 4K) Pa vector = (x1, Y1 + 4K) = k1af vector = K1 (4-x1, - Y1) Pb vector = (X2, Y2 + 4K) = k2bf vector = K2 (4-x2, - Y2), so K1 + K2 = X1 / (4-x1) + x2 / (4-x2) = (4 (...)

The solution of x2 + 3x-1 = 0

5X-1=0
5X=1
X=0.2

In the parallelepiped abcd-a'b'c'd ', we know that the volume diagonal a'c = 4, b'd = 2. If P is a point in space and PA' = 3, PC = 5, then PB '^ 2 + PD ^ 2 =?

Here I am
Pa'a'c and PC are 3, 4 and 5, respectively
So PA 'is perpendicular to a'c
Let the intersection of a'c and b'd be o
So o divides a'c and b'd equally
So a'o = 2
At rtpa'o
Po = 13 ^ 0.5 (root 13)
PB'^2=PO^2+B'O^2-2PO*B'O*CosPOB'
PD^2=PO^2+DO^2+2PO*DO*CosPOB'
So PB '^ 2 + PD ^ 2 = 2PO ^ 2 + do ^ 2 + b'o ^ 2 = 13 * 2 + 1 + 1 = 28

The square of 7x-x = 6 x =?

7x-x^2=6 x^2-7x+6=0 (x-1)(x-6)=0 x1=1 x2=6

Set a = {(x, y) y = - X & # 178; + MX-1}, B = {(x, y) y = 3-x, 0 ≤ x ≤ 3}. If a ∩ B is a set with only one element, the value range of real number m is obtained

Simultaneous 3-x = x ^ 2 + MX-1
x^2+(m+1)x-4=0
It is to find the point of quadratic equation with one variable and only one root
delta=(m+1)^2-4*1*4=0
（m+1）^2=16
M = 3 or M = - 5

(play an idiom) 0 + 0 =?, 1 * 1 = 1,7811001000 * 1000 = 100 * 100 * 100

0 + 0 =?, --- nothing
1 * 1 = 1, -- invariable
78, -- seven up and eight down
1 / 100 --- one in a hundred
1000 * 1000 = 100 * 100 * 100 --- do everything possible

If a parallelogram with a base of 7cm and a height of 5cm is drawn into a rectangle, the area will be increased by 12 square centimeters, and the perimeter of the rectangle is cm
It's a process

7 * 5 + 12 = 49 (area of rectangle)
49 / 7 = 7 (height of rectangle)
So, perimeter = 7 * 2 + 7 * 2 = 28

Solving equation 40% x + 1 / 12 = 58

40%x+1\12=58
4.8x+1=696
4.8x=695
x=3475/24

Find the tangent equation and normal equation at the point M0 on the curve y = f (x) [f (x) = x & # 178; 1 / 2, M0 (1,1)]

Alas! Can't you express "1 / x ^ 2" as "1 / x ^ 2"? It's very harmful! F '(x) = - 2x ^ (- 3) k-cut = f' (xm0) = - 2 * 1 ^ (- 3) = - 2 k-method = - 1 / k-cut = 1 / 2  tangent equation y-ym0 = k-cut (x-xm0) = > Y-1 = - 2 (x-1) = > 2x + Y-3 = 0 normal equation y

How to solve 7x + x = 42.8

7x+X=42.8
8X=42.8
X=5.35