This is a math problem of Liberal Arts in senior high school The elliptic equation x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 (a > b > 0), a focus of which coincides with the focus of parabola y ^ 2 = 8x, the eccentricity e = 2 √ 5 / 5, the right focus f passing through the ellipse is a straight line L which is not perpendicular to the coordinate axis, intersecting the ellipse at a and B (2) Let point m (1,0), and (vector Ma + vector MB) be perpendicular to vector AB, find the equation of line L Please help explain the steps, thank you very much_ ∩)O~

Parabola focus (2,0) ellipse C = 2. C ratio a = e = 2 √ 5 / 5. A = √ 5. B = 1, the equation is X & sup2. / 5 + Y & sup2; = 1. Let y = K (X-2) ② simultaneous ① ② → (5K & sup2; + 1) x & sup2; - 20K & sup2; X + 20K & sup2; - 5 = 0. ③ get X1 + x2 = 20K & sup2; / (5K & sup2; + 1), Y1 + y2 = - 4K / (5K & sup2; + 1)

The equation (m-2) x & # 178; - 3mx + 6m-5 = 0 is a quadratic equation with one variable, then M satisfies the condition that?

36x ^ 2 = 1 x ^ 2 = 1 / 36 x = plus or minus one sixth: 4x ^ 2 = 81

Help me solve the equation 95 + 5% x = x 700-40% x = x 60-20% x = x

Solving equation 95 + 5% x = x
95=x-5%x
95=0.95x
x=95/0.95=100
700-40%x=x
700=40%x+x
700=1.4x
x=700/1.4=500
60-20%x=x
60=20%x+x
60=1.2x
x=60/1.2=50

Solve the equation x & # 178; = X

X & # 178; - x = 0, X (x-1) = 0, x = 0 or x = 1

(17-y) &# 178; + Y & # 178; = 169, solve the equation,

(17-y）²+y²=169
2y²-34y+289=169
y²-17y+60=0
(y-12)(y-5)=0
y1=12 y2=5

How to solve X & # 178; + Y & # 178; = 10 & # 178; (x + 9) &# 178; + Y & # 178; = 17 & # 178;

x²+y²=10²
（x+9）²+y²=17²
How about the intersection of two circles?
If you follow the rules of solving the equation
The second formula subtracts the first formula to get
(x+9-x)(x+9+x)=(17+10)(17-10)
9(2x+9)=27*7
2x+9=21
2x=12
x=6
Y = plus or minus 8

121(x－y）²－169(x＋y）

Original formula = [11 (X-Y) + 13 (x + y)] [11 (X-Y) - 13 (x + y)]
=(24x-2y)(-2x-24y)
=-4(12x-y)(x+12y)
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If you have any new questions, please don't send them in the form of follow-up questions, send them to me for help or send them to the question link address,

If (x + Y-5) &# 178; + x-3y-17 = 0, find the value of X-Y

If the sum of two non negative numbers is 0, the sum of them can only be 0 if they are both 0
(x + Y-5) &# 178; + x-3y-17 = 0
(x + Y-5) & #178; > = 0, x-3y-17 > = 0
(x + Y-5) ² = x-3y-17 = 0
x＋y-5=0,x-3y-17=0
The solution is: x = 8, y = - 3
x-y=11

Given - x + 3Y = 5, then the value of 5 (x-3y) 2-8 (x-3y) - 5 is ()
A. 80B. -170C. 160D. 60

∵ - x + 3Y = 5, ∵ x-3y = - 5, ∵ 5 (x-3y) 2-8 (x-3y) - 5 = 5 × (- 5) 2-8 × (- 5) - 5 = 125 + 40-5 = 160

Verification: √ X & # 178; + Y & # 178; > &# 179; √ X & # 179; + Y & # 179; how to prove? Please give the steps,

If x ^ 3, y ^ 3 = 0,
Both sides at the same time to the sixth power get: (x ^ 2 y ^ 2) ^ 3 > = (x ^ 2 y ^ 2)