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Find the coordinates of the intersection point of the image of the function y = 2x + 1 and y = - 3x-4
If the first equation is reduced by the second one, we get 5x + 5 = 0
x=-1,y=-1
（-1，1）
（-1，-1）
Solve the equations
y=2x+1
y=-3x-4
X = - 1, y = - 1
Intersection coordinates: (- 1, - 1)
Since it is an intersection point, it satisfies two functions at the same time
2x+1=-3x-4
5x=-5
x=-1
Substituting any function, y = 2 * (- 1) + 1 = - 1
So the intersection coordinates (- 1, - 1)
It is known that the focal length of the ellipse x2a2 + y2b2 = 1 (a > b > 0) is 2c, and a, B, C form an arithmetic sequence in turn, then the eccentricity of the ellipse is ⊙___ .
∵ a, B, C in turn into a sequence of arithmetic numbers, ∵ 2B = a + C, A2-B2 = C2, ∵ a2 - (a + C2) 2 = C2, namely & nbsp; 3a2-5c2-2ac = 0, ∵ 5e2-2e + 3 = 0, e = 35 & nbsp; or & nbsp; e = - 1 (rounding off)
Given that the image of the first-order function y = 3x + m and y = - 2X-4 intersects at a point in the third quadrant, then the value range of M is
-4
Find intersection 3x + M = - 2X-4 5x = - M-4 x = - (M + 4) / 5
y=(2m-12)/5
Because it's in the third quadrant, so X
The eccentricity of an ellipse is ()
A. 45B. 35C. 25D. 15
Let the major axis be 2a, the minor axis be 2B, and the focal length be 2c, then 2A + 2C = 2 × 2B, that is, a + C = 2B {(a + C) 2 = 4B2 = 4 (a2-c2), so 3a2-5c2 = 2Ac, divide A2, and get 5e2 + 2e-3 = 0, ∧ e = 35 or E = - 1 (rounding off), so select B
If the image with the function f (x) = 1 / 3x ^ 2 + 1 / 2x ^ 2-2x + m does not pass through the fourth quadrant, then the value range of M is
First, merge the similar terms, the original function = 5x ^ 2 / 6-2x + m classification discussion
(1) When the discriminant is less than or equal to 0, the image is greater than or equal to 0, and must not pass through the fourth quadrant
The discriminant = 4-10m / 3 is less than or equal to 0, and the solution m is greater than or equal to 6 / 5
(2) When the discriminant is greater than 0, we must ensure that the two components of the function are less than 0
Two less than 0, the most intuitive expression is
X1x2 greater than 0
X1 + x2 less than 0
But according to Weida's theorem, X1 + x2 = 12 / 5 is greater than 0, so there is no solution
In conclusion, M is greater than or equal to 6 / 5
The range of M is [5 / 6, positive infinity]
If the length of the major axis, the length of the minor axis and the focal distance of an ellipse form an arithmetic sequence, then the eccentricity is zero
The answer is three fifths... But I can't figure it out... Baidu uses a, B, C to form the arithmetic sequence... But the long axis is not 2a, the short axis is not 2B, why use the half axis to calculate
If 2a, 2b and 2c are equal, then:
4b=2a＋2c
2b=a＋c
4b²=(a＋c)²
4(a²－c²)=a²＋2ac＋c²
Namely:
3a²－2ac－5c²=0
(a＋c)(3a－5c)=0
Then:
3a－5c=0
e=c/a=3/5
It's 2a, 2b, 2C
So 4B = 2A + 2C
So 2B = a + C
It is equivalent to ABC
It's OK to use 2a2b2c, just like ABC! The first thing to do is to be flexible. Don't complicate simple problems!
If the image of function y = 2x power + m does not pass through the second quadrant, then the range of M is
If the image of function y = 2x power + m does not pass through the second quadrant, then the range of M is
[solution]: y = 2 ^ x + M
M
The elliptic equation x ＾ 2 / 2C ＾ 2 + y ＾ 2 / C ＾ 2 = 1, e = √ 2 / 2, there are two symmetric points m (2,1) on the ellipse, and the range of focal length is calculated
Let a point be a (XA, ya), then the symmetric point be B (4-xa, 2-ya). A and B are both on the ellipse: XA ^ 2 + 2ya ^ 2 = 2C ^ 2xa ^ 2-8xa + 16 + Ya ^ 2-4ya + 4 = 2C ^ 2 subtraction: 8xa + 4ya-20 = 0, so 2x + Y-5 = 0 is the line where AB is. Substitute y = - 2x + 5 into the elliptic equation: x ^ 2 + 2 (- 2x + 5) ^ 2 = 2C ^ 29x ^ 2-40x + 50
The focal length is C, M is in the ellipse, and the ellipse becomes x ^ 2 + 2Y ^ 2 = 2C ^ 2
X^2+2Y^2>2^2+2*1^2=6
C> Radical (3)
Given the function y = (M + 1) x + 2m-6, the image of this function is parallel to the straight line y = 2x + 5, the analytic expression of the function is obtained
If two lines are parallel, the slope k is equal, y = KX + B
That is: M + 1 = 2 get: M = 1
So 2m-6 = - 4
So the analytic expression of the function is y = 2X-4
m+1=2.m=1,2m-6=-4.
The analytical formula is y = 2X-4
Because the two lines are parallel, so m + 1 = 2, that is, M = 1, so y = 2X-4
1) If the focal length of the ellipse x ^ 2 / 4 + y ^ 2 / M = 1 is 2, then M=
2) Given that the coordinates of two vertices of △ ABC are a (- 4,0), B (4,0), and the perimeter of △ ABC is 18, the trajectory equation of vertex C is obtained
3) Which is flatter, ellipse 9x ^ 2 + 4Y ^ 2 = 36 or ellipse x ^ 2 / 25 + y ^ 2 / 16 = 1?
4) If a vertex and two focal points of an ellipse form an equilateral Delta, what is the eccentricity of the ellipse?
5) It is known that the symmetry axis of the ellipse is the coordinate axis, 0 is the coordinate origin, f is a focus, a is a vertex. If the length of the major axis of the ellipse is 26, and the cos angle ofa = 5 / 13, the standard equation of the ellipse is obtained
Today, I fell asleep in math class and didn't listen at all. How can I judge whether the long axis of an ellipse is on the X axis or the Y axis? Is the smaller the eccentricity, the rounder the ellipse?
1) M = 4 or 5
2) The trajectory equation of vertex C is x ^ 2 / 25 + y ^ 2 / 9 = 1
3) Eccentricity of ellipse 9x ^ 2 + 4Y ^ 2 = 36 e = SQR (5) / 3
Eccentricity of ellipse x ^ 2 / 25 + y ^ 2 / 16 = 1 E = 3 / 5
The bigger the eccentricity is, the flatter the ellipse is, so the former is more flatter
4) A = 2c, so e = C / a = 1 / 2
5) Because 2A = 26, so a = 13
Cos angle ofa = 5 / 13 = C / A,
So, C = 5
b^2=a^2-c^2=144.
The standard equation of this ellipse is: x ^ 2 / 169 + y ^ 2 / 144 = 1 (focus on X axis) or x ^ 2 / 144 + y ^ 2 / 169 = 1 (focus on Y axis)
High school sophomore! Don't sleep any more
To judge whether the major axis of an ellipse is on the X or Y axis, it's very easy to look at x ^ 2; the denominator of Y ^ 2 is the one with the largest denominator, that is, the axis where the focus is (of course, we need to change it to the standard form before we look at it);
The eccentricity of ellipse is 0
yes
I will know how to do it if I use my brain!!! I'm getting worse in math now, only 90 points!! alas
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