If the common focus of ellipse x26 + Y22 = 1 and hyperbola x23-y2 = 1 is F1, F2, P is an intersection of two curves, then the value of cos ∠ f1pf2 is () A. 13B. 23C. 73D. 14

If the common focus of ellipse x26 + Y22 = 1 and hyperbola x23-y2 = 1 is F1, F2, P is an intersection of two curves, then the value of cos ∠ f1pf2 is () A. 13B. 23C. 73D. 14

Let p be a point on the right branch of hyperbola, let | Pf1 | = m, | PF2 | = n. then M-N = 23m + n = 26, the solution is Mn = 3. | F1F2 | = 4. | cos ∠ f1pf2 = M2 + n2-422mn = (M + n) 2-2mn-422mn = 24-6-162 × 3 = 13
If the ellipse x ^ 2 / M + y ^ 2 / N = 1 and the hyperbola x ^ 2 / a-y ^ 2 / b = 1 have the same focus, F1, F2, P are an intersection of two straight lines
Then the value of | Pf1 | * | PF2 | is
The sum of the distances from any point P of the ellipse to the two focal points is 2 √ M,
Pf1 + PF2 = 2 √ m,
The absolute value of the distance difference between any point of hyperbola and two focal points is 2 √ a,
That is | pf1-pf2 | = 2 √ a, the square difference of the above two formulas is 4pf1 * PF2 = 4m-4a
Therefore, Pf1 * PF2 = m-a
If the common focus of ellipse x26 + Y22 = 1 and hyperbola x23-y2 = 1 is F1, F2, P is an intersection of two curves, then the value of cos ∠ f1pf2 is ()
A. 13B. 23C. 73D. 14
Let p be a point on the right branch of hyperbola, let | Pf1 | = m, | PF2 | = n. then M-N = 23m + n = 26, the solution is Mn = 3. | F1F2 | = 4. | cos ∠ f1pf2 = M2 + n2-422mn = (M + n) 2-2mn-422mn = 24-6-162 × 3 = 13
If the solution of the equation AX + 199 = 3x + 200 about X is x = 1, then the value of 2a-1 is
X=1
Substituting
a+199=3+200
A=4
2a-1=8-1=7
[x(x^2y^2-xy)-y(x^2-x^3y)]/3x^2y
=(x^3y^2-x^2y-x^2y+x^3y^2)/3x^2y
=(2x^3y^2-2x^2y)/3x^2y
=2x^3y^2/3x^2y-2x^2y/3x^2y
=(2xy-2)/3
Application of square root
① For the values of \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\forany nonnegative number a, (√ a) &# 178; what is equal to?
So, we're going to find out 2 and \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ = 49, (√ 0) & 178
The equation (a + 2) · X & # 178; - 2aX + a = 0 of X has two unequal real roots
The equation (a + 2) · X & # 178; - 2aX + a = 0 of X has two unequal real roots, which are X1 and X2 respectively. The two intersections of parabola y = x & # 178; - (2a + 1) x + 2a-5 and X axis are on both sides of point (2,0)
(1) Finding the value range of real number a
(2) When | x1 | + | x2 | = 2 √ 2, find the value of A. {the process needs to use WIDA's theorem}
According to: two unequal real number roots, get: a
（x+4）×（x+2）=x²+56
x2+2X+4X+8=x2+56
6X=56-8
X=48/6
X=8
Square root problem
A field is 100 meters to 110 meters wide, 64-75 meters wide. There is a rectangular field, which is 1.5 times as wide and covers an area of 7560 square meters. Can this field meet the standard of that field?
This question, thought for a long time!
64*1.5=96
75*1.5=
1.5x^2=7560
x=3040
About 50
60*1.5=90
Obviously, it can't meet the requirements
The two real roots of the equation x ^ 2-2ax + 9 = 0 of X are α and β respectively. Find the minimum value of (α - 1) ^ 2 + (β - 1) ^ 2
Delta = 4A ^ 2-36 > = 0, a > = 3 or a = 3, or a
According to Weida's theorem:
αβ=9，,α+β=2a
(α-1)^2+(β-1)^2
=α^2-2α+1+β^2-2β+1
=(α+β)^2-2αβ-2(α+β)+2
=4a^2-4a-16
=4(a-1/2)^2-17
Also: 4A ^ 2-36 0, | a | 3
When a = 3, 4 (A-1 / 2) & # - 17 = 8 is the smallest
α+β=a；
αβ=9；
（α-1)^2+(β-1)^2
=(α+β)^2-2αβ-2(α+β)+2
=a^2-18-2a+2
=a^2-2a-16
The answer is 8
The two real roots of the equation β ^ 2aX + 9 are respectively,
∴α＋β＝﹣﹙﹣2a﹚＝2a
αβ＝9
∴ （α-1)^2+(β-1)^2＝﹙α²－2α＋1﹚＋﹙β²－2β＋1﹚
＝﹙α²＋β²﹚－2﹙α＋β﹚＋2
4*2.5+4x=56
10＋4x＝56
4x＝56－10
4x＝46
x＝11.5
4*2.5+4x=56
That is, 10 + 4x = 56
That is, 10 + 4x-10 = 56-10
That is 4x = 46
That is, x = 23 / 2
4*2.5+4x=56
10+4x=56
4x=56-10
4x=46
x=46/4
x=23/2
I don't understand. Please ask.