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The eccentricity of the ellipse with focus on the x-axis is two-thirds root sign three, and the ellipse intersects with (X-2) ^ + (Y-1) ^ = 5 / 2 at two points a and B. the length of line AB is equal to the diameter of the circle 1. Find the equation of line ab 2. Find the equation of ellipse
1. Let e = √ 3 / 2, so C & sup2; = 0.75a & sup2;, so B & sup2; = 0.25A & sup2;, so let the ellipse be X & sup2; + 4Y & sup2; = A & sup2;, because the length of the line AB is equal to the diameter of the circle, so the line AB must pass P (2,1). Let the line AB be Y-1 = K (X-2), Y-1 = K (X-2) and X & sup2; + 4Y & su
GGF
Given that the focus of the ellipse where the square of X divided by the square of T plus the square of Y divided by 5T equals 1 is (0, positive and negative root sign 6), then the real number T equals?
x^2/t^2+y^2/5t=1
a^2=5t,b^2=t^2,c^2=5t-t^2
C = root 6
That is, 5t-t ^ 2 = 6
t^2-5t+6=0
（t-2)(t-3)=0
Because there is a > b, that is 5T > T ^ 2, t
(5t)^2-t^2=6
24t^2=6
t=1/2
T equals 6
If the quadratic + 4 of the function y = 6x-b is a positive proportional function, the value of B can be obtained. (2) if the function y = (A-2) x + (3 + a) is a positive proportional function,
(3) if the image of the square of a of positive scale function y = (2-3a) x passes through two or four quadrants, the value of a is calculated
(1) Because the function is a positive proportion function, so - B ^ 2 + 4 = 0. So B = 2 or - 2
(2) Similarly = > 3 + a = 0, so a = - 3. The function y = - 5x is obviously a monotonic decreasing function, so y decreases with the increase of X
(3) Obviously, a ^ 2 = 1, so a = 1 or - 1, and because the positive proportion function passing through two or four quadrants is a decreasing function, when a = - 1, y = 5x. It is not a decreasing function. When a = 1, y = - x is a decreasing function, so a = 1
How to draw this image?
This is a simple piecewise function, divided into two cases, when x is greater than or equal to 0, y = X. conversely, y = - x, respectively draw two different cases of the image
How to draw the image of absolute value function?
Using symmetry
How to draw the image of quadratic function with absolute value?
If the absolute value is added to x, the left part of the y-axis image of the original function without absolute value can be rounded off, and the right part can be folded; if the absolute value is added to the whole, the negative half axis image of the y-axis image of the original function without absolute value can be turned down, and then the lower part can be rounded off. Or the point drawing method can be used directly
It is known that Tana = - 1 / 3, and calculated
(Sina + 2cosa) / (5cosa Sina) I don't have any points, please help me to thank this ~ process ~ tomorrow, thank you 1
The numerator and denominator are divided by cosa
The original formula = (Sina / cosa + 2) / (5-sina / COSA)
=(tana+2)/(5-tana)
=5/16
Image of absolute value of quadratic function
The absolute value is outside and inside
If the absolute value is outside, turn all the parts of the function below the x-axis to the x-axis
If the absolute value is in it, first delete all the parts on the left side of the Y-axis of the function, and then turn the part on the right side of the y-axis to the left side of the y-axis
1. Given α∈ (π / 2, π) and sin α = 1 / 2, if sin (α - β) = - 3 / 5, β∈ (π / 2, π), find the value of cos β
∵sinα=1/2
∴α=150°
∴α-β∈〔-30°,60°〕
∴cos(α-β)= 4/5
∴cos(150°-β)=√3/2cosβ+1/2sinβ=-3/5
sin(150°-β)=1/2cosβ-√3/2sinβ=4/5
Can solve cos β =... Solve it yourself
How to draw the image of quadratic function with absolute value
y=|x^2-x-6|-2x
You can simplify it first and turn it into a piecewise function
When - ∞ < x < - 2, y = x & sup2; - 3x-6
When - 2 ≤ x ≤ 3, y = - X & sup2; - x + 6
When x > 3, y = x & sup2; - 3x-6
And then draw the image according to the segmentation function
RELATED INFORMATIONS
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