According to the conditions, the standard equation can be solved. Given that one focus of the ellipse is (0,2) and the eccentricity is 1 / 2, the standard equation can be solved

According to the conditions, the standard equation can be solved. Given that one focus of the ellipse is (0,2) and the eccentricity is 1 / 2, the standard equation can be solved

One focus is (0,2)
Explain that the focus is on the y-axis
C=2
Eccentricity e = C / a = 1 / 2
∴a=4
a²=16
b²=a²-c²=16-4=12
The elliptic standard equation is
y²/16+x²/12=1
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It is known that the coordinates of the two focal points of an ellipse are (- 2,0) (2,0) respectively, and one of its vertex coordinates is the standard equation of (0,3)
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The focus is on the x-axis, so: C = 2, B = 3, a ^ 2 = 2 ^ 2 + 3 ^ 2 = 13
So the ellipse is: x ^ 2 / 13 + y ^ 2 / 9 = 0
The focus is on the x-axis, C = 2, B = 3, a ^ 2 = 2 ^ 2 + 3 ^ 2 = 13
The ellipse is: x ^ 2 / 13 + y ^ 2 / 9 = 0
The elliptic standard equation passing through point (3, - 2) with a focus of (- √ 5,0)?
The root of the equation Q ^ 2 and Q ^ 2 is 0
If the sum of abscissa of the intersection of the straight line y = KX + 1 and the curve X ^ 2 + y ^ 2 + x-ky = 0 is zero, find the real number K
By substituting y = KX + 1 into the equation x ^ 2 + y ^ 2 + x-ky = 0, we can simplify x ^ 2 + (KX + 1) ^ 2 + x-k (KX + 1) = 0, and obtain that the sum of abscissa of intersection points of (k ^ 2 + 1) x ^ 2 - (k ^ 2-2k-1) x + (1-k) = 0 is 0, that is, X1 + x2 = 0
It's not easy.
It is known that ab belongs to (0, Pai / 2) and tanb = sinacosa / 1 + sin ^ 2A. 1
It is known that ab belongs to (0, Pai / 2) and tanb = sinacosa / 1 + sin ^ 2A
1. To express tanb as Tana
2 find the maximum value of tanb, and find the value of Tan (a + b) when tanb reaches the maximum value
(1) Tanb = sinacosa / (COS ^ 2 A + 2 · sin ^ 2 a) = Tana / (1 + 2 Tan ^ 2 a) (divide the denominator by cos ^ 2 a at the same time) (2) let u = Tana, then u ＞ 0 tanb = u / (1 + 2 u ^ 2) = 1 / (2U + 1 / U) ∵ 2U + 1 / u ≥ 2 · radical (2U · 1 / U) = 2 · radical 2 (equal sign if and only if u = radical 2 / 2)
In the plane rectangular coordinate system, the two vertices a and C of the square oabc with side length 2 are respectively on the positive half axis of Y axis and X axis,
(1) Area = OA * OA * 3.14 * 45 / 360 = 1.57 (2) when Mn and AC are parallel, am / AB = CN / CB, because AB = CB, am = CN, △ OAM ≌ △ OCN ∠ AOM = ∠ con and ∠ con = ∠ yoa (because of simultaneous rotation), so ∠ yoa = 22.5 ° because of AB = CB
If the line y = KX is the tangent at a point on the curve y = x3-3x2 + 2x, then the real number k=______ .
The derivative of Y ′ = 3x2-3x2-3x2 + 2x is y ′ = 3x2-6x + 2 set the tangent point coordinate is (x0, Y0) the slope of the tangent is k = 3x02-6x0 + 2, the tangent equation is y-y0 = (3x02-6x02-6x00 + 2) (x-x02-3x2-3x2-3x2-3x2-3x2-6x + 2x2, the tangent point coordinate is (x0, Y0, Y0) the point coordinate is (X00, Y0, Y0) the tangent equation is y equation is y = (3x02-3x02-6x02-6x02-6x02-6x00 + 2) x - (3x02-6x02-6x02-6x02-6x02-6x02-6x00-6x00-6x00 + 2-6x00 + 2 + 2 + 2 + 2 + 2 + 2 + 2) X00 (X00 (x00-3x0-‖ k = 2 Or - 14, so the answer is 2 or − 14
  known as 0 ＜ B ＜ quarter faction, quarter faction ＜ a ＜ three quarter faction, cos (quarter faction + a) = - 5 / 3, sin (three quarter faction + b) = 13 / 5
∵0
In the plane rectangular coordinate system, the two vertices a and C of the square oabc with side length 2 are respectively on the positive half axis of Y axis and X axis, and the point O is at the origin. Now rotate the square oabc clockwise around the point O, and stop rotating when the point a falls on the line y = x for the first time. In the process of rotation, AB intersects the line y = x at the point m, BC intersects the X axis at the point n (as shown in the figure). (1) calculate the direction swept by the side OA in the process of rotation (2) in the process of rotation, when Mn and AC are parallel, calculate the degree of rotation of the square oabc; (3) if the perimeter of △ MBN is p, does the value of P change in the process of rotating the square oabc? Please prove your conclusion
(1) The angle between the line y = x and the Y axis is 45 ° and the OA rotates 45 ° for the first time. The area swept by OA during the rotation is 45 π × 22360 = π 2. (2) ∵ Mn ∥ AC, ∵ BMN = ∵ BAC = 45 ° and ∵ BNM = ∵ BCA = 45 ° In the rotation process, when Mn and AC are parallel, the rotation degree of oabc is 45-22.5 ° = 22.5 °. (3) there is no change of P value in the rotation process of oabc. It is proved that when the Y-axis of Ba intersection is extended to e point, ∠ AOE = 45 ° - AOM, ∠ con = 90 ° - 45 ° - AOM = 45 ° - AOM, ∠ AOE = con C. In the process of rotating the square oabc, the value of P does not change
Given that the curve X-Y ^ 2-1 = 0 intersects with the straight line kx-y = 0, find the value range of the real number K y = KX and substitute it into x-k & # 178; X & # 178; - 1 = 0, K & # 178; X & # 178; - x + 1 = 0, that is, if the equation has a solution k = 0, then - x + 1 = 0, and K ≠ 0, then the discriminant △ = 1-4k & # 178; ≥ 0 (how does the discriminant come from, why is it written like this)
There are intersection points between straight line and curve, so equations are needed
｛x-y^2-1=0
｛y=kx
There are real solutions
= = >
x-k²x²-1=0
k²x²-x+1=0
When k = 0,
The equation is not quadratic, the equation is - x + 1 = 0 and has real solution;
When k ≠ 0,
The equation is a quadratic equation of one variable, and the condition of real number solution is
△=1-4k²≥0
Quadratic equation AX & # 178; + BX + C = 0 (a ≠ 0)
The condition of real number solution is △ = B & # 178; - 4ac ≥ 0
The discriminant △ = 1-4k & # 178; ≥ 0
The discriminant △ = square B - 4ac ≥ 0 means there are two roots.