It is known that the straight line L: x + y-m = 0 (m ＞ 0), the ellipse x squared divided by a and Y squared divided by B is equal to 1, Verification: the distance from the intersection of X and Y axes is greater than Greater than a + B

It is known that the straight line L: x + y-m = 0 (m ＞ 0), the ellipse x squared divided by a and Y squared divided by B is equal to 1, Verification: the distance from the intersection of X and Y axes is greater than Greater than a + B

What is the condition that M + 2 / x square + 1-m / y square equals 1 is an ellipse,
(m+2)(1-m)>0
-2
It is known that the square of a quarter of x plus the square of y of an ellipse G is equal to the tangent l intersection ellipse of a circle x plus the square of Y through a point (m, 0)
AB two points (1) focus coordinates and eccentricity of spherical ellipse g (2) express the absolute value of AB as a function of M and find the maximum value of absolute value of ab
1)a^2=4,b^2=1,c^2=a^2-b^2=3,
(0,3), √ is the focus,
Centrifugation e = C / a = √ 3 / 2
2) The center of the circle x ^ 2 + y ^ 2 = 1 is (0,0), the radius r = 1, obviously | m | > 1
Let the tangent equation of a circle passing through M (m, 0) be y = K (x-m),
Because the distance from the center of the circle to the tangent is r = 1,
So from | km | / √ (k ^ 2 + 1) = 1, the solution is k ^ 2 = 1 / (m ^ 2-1), (1)
Substituting the tangent equation into the elliptic equation, we get x ^ 2 + 4 [K (x-m)] ^ 2 = 4,
It is reduced to (4K ^ 2 + 1) x ^ 2-8k ^ 2mx + 4K ^ 2m ^ 2-4 = 0,
Let a (x1, Y1), B (X2, Y2),
Then X1 + x2 = 8K ^ 2m / (4K ^ 2 + 1), X1 * x2 = (4K ^ 2m ^ 2-4) / (4K ^ 2 + 1),
So | ab | ^ 2 = (x2-x1) ^ 2 + (y2-y1) ^ 2 = (1 + K ^ 2) [(x1 + x2) ^ 2-4x1 * x2] =
Let a > 1, then the range of eccentricity e of hyperbola x2a2 − Y2 (a + 1) 2 = 1 is ()
A. (2，2)B. (2，5)C. （2，5）D. (2，5)
E2 = (CA) 2 = A2 + (a + 1) 2A2 = 1 + (1 + 1a) 2, because 1a is a decreasing function, when a > 1, 0 < 1A < 1, so 2 < E2 < 5, that is, 2 < e < 5, so B
Known - (yuan / 2)
The root formula is Tana + tanb = - 6
tana×tanb=7
So tan (a + b) = (Tana + tanb) / (1-tana × tanb) = 1
So a + B = 45 degrees
45 degrees. Trust me. It's a long time
π. It's PI, but the others, I don't know. I'm only in Grade 6
tana+tanb=-6
tana×tanb=7
tan(a+b)=(tana+tanb)/(1-tana×tanb)=1
So a + B = 45 degrees
Given the positive proportional function y = KX, when x increases by 3, y decreases by 2
According to the problem: y = KX, Y-2 = K (x + 3)
Then the point (- 3,2) is on the function
Then k = - 2 / 3
Let a > 1, then the range of eccentricity e of hyperbola x2a2 − Y2 (a + 1) 2 = 1 is ()
A. (2，2)B. (2，5)C. （2，5）D. (2，5)
E2 = (CA) 2 = A2 + (a + 1) 2A2 = 1 + (1 + 1a) 2, because 1a is a decreasing function, when a > 1, 0 < 1A < 1, so 2 < E2 < 5, that is, 2 < e < 5, so B
Known negative half
tana+tanb=-60,
So Tana and tanb are negative,
So the negative half
If we know the positive scaling function y = KX, when x = - 2, y = 6, then the scaling coefficient is
Scale factor k = 6 / (- 2) = - 3
Directly substituting y = 6 and x = - 2 into the solution, we get k = - 3, that is, the scale coefficient is - 3
We know the equation x & # 178; - (M + 2) x + 2m = 0 (1) and prove that the equation always has two unequal real roots (2) if this is the case
We know the equation x & # 178; - (M + 2) x + 2m = 0 (1) prove that the equation always has two unequal real roots (2) if one root of the equation is 1, ask for the other root of the equation, and find the perimeter of the right triangle with these two sides
The equation x & # 178; - (M + 2) x + 2m = 0 for X is known
(1) To prove that the equation always has two unequal real roots
x²-(m+2)x+2m=0
△=[-(m+2)]²-4*2m=m²+4m+4-8m=m²-4m+4=(m-2)²≥0
(2) If one root of the equation is 1, ask for the other root of the equation, and find the perimeter of the right triangle with these two sides
When m = 2, the original equation has only one root
First, I think there is a problem
First method:
x²-(m+2)x+2m=0
（x－m）（x－2）＝0
When m = 2, there is only one real root
Second method:
b²－4ac
＝(m+2)²-8m
=m²-4m+4
=(m-2)²>=0
Second question:
When x = 1, X & # 178; - (M + 2) x +... Expansion
First, I think there is a problem
First method:
x²-(m+2)x+2m=0
（x－m）（x－2）＝0
When m = 2, there is only one real root
Second method:
b²－4ac
＝(m+2)²-8m
=m²-4m+4
=(m-2)²>=0
Second question:
When x = 1, X & # 178; - (M + 2) x + 2m = 1-m-2 + 2 = 0
M=1
The equation is
x²-3x+2=0
（x－2）（x-1)=0
X = 1 or x = 2
When these two roots are right angles, the root is, the oblique side length = root 5, the perimeter = 3 + root 5
When 2 is a hypotenuse, the other right angle side = radical 3, perimeter = 3 + radical 3, and fold up
Find Delta, delta > 0, so there are two different solutions. When a solution is 1, substitute 1, calculate m, and then find the second solution
（1）△=(m+2)² -4*2m
=m² +4m+4-8m
=m² -4m+4
=(m-2)² ≥0
When m = 2, the equation has two equal real roots
When m is not equal to 2, the equation has two unequal real roots.
(2) If one of the roots of the equation is 1, the equation is brought in
1 - (M + 2)... Expansion
（1）△=(m+2)² -4*2m
=m² +4m+4-8m
=m² -4m+4
=(m-2)² ≥0
When m = 2, the equation has two equal real roots
When m is not equal to 2, the equation has two unequal real roots.
(2) If one of the roots of the equation is 1, the equation is brought in
1-（m+2)+2m=0
So m = 1
Bring in the original equation x & # 178; - 3x + 2 = 0
So the solution is: x = 1 or x = 2
Two are right triangle sides
So the third side = √ (1 + 4) = √ 5
Perimeter = 1 + 2 + √ 5 = 3 + √ 5