Find the elliptic equation which passes through the point (- 3 / 2,5 / 2) and 9x ^ 2 + 4Y ^ 2 = 45 and has a common focus with the ellipse

Find the elliptic equation which passes through the point (- 3 / 2,5 / 2) and 9x ^ 2 + 4Y ^ 2 = 45 and has a common focus with the ellipse

Firstly, it is transformed into the standard form X ^ 2 / 5 + y ^ 2 / (45 / 4) = 1, C ^ 2 = 45 / 4-5 = 25 / 4, and its focus is F1 (0,5 / 2); F2 (0, - 5 / 2) passes through point a (- 3 / 2,5 / 2), so the sum of the distances from a to F1 and F2 is the fixed length AF1 = √ (9 / 4) = 3 / 2af2 = √ ((9 / 4) + 100 / 4) = √ 109 / 2; 2A = AF1 + af2 to get a
Given that the center of the ellipse is at the origin, the eccentricity is equal to 1 / 2, and one focus of the ellipse coincides with the focus of the parabola y = - 4x, then the elliptic equation is?
Because the eccentricity is 1 / 2, so a = b = 2C! The focus of parabola coincides, so a = 2! So the answer comes!
The focus of E = C / a = 1 / 2 parabola y = - 4x is - P / 2 = 1 C = 1 a = 2 b = root 3, so the equation is x ^ 2 / 4 + y ^ 2 / 3 = 1. If you have any questions, you can ask! Forget to adopt! thank you
It is known that the center of the ellipse is at the coordinate origin o, a focus coincides with the intersection of the parabola y ^ 2 = 4x, and the eccentricity of the ellipse is √ 2 / 2,
When the area of △ AOB is the largest, it is best to give the answer within two hours
Parabola: y ^ 2 = 2 * 2x, focus f (1,0),
For ellipses, C = 1, e = C / a = √ 2 / 2, a = √ 2,
b^2=a^2-c^2=2-1=1,
The elliptic equation is: x ^ 2 / 2 + y ^ 2 = 1
2. Let the linear equation be y = KX + 2, a (x1, Y1), B (X2, Y2),
The distance between O and ab is d = 2 / √ (1 + K ^ 2),
x^2/2+(kx+2)^2=1,
(1+2k^2)x^2+8kx+6=0,
According to Veda's theorem,
x1+x2=-8k/(1+2k^2),
x1*x2=6/(1+2k^2),
According to the chord length formula:
|AB|=√（1+k^2）(x1-x2)^2=√（1+k^2）[(x1+x2)^2-4x1x2]
=√（1+k^2）[64k^2/(1+2k^2)^2-24/(1+2k^2)]
=√（1+k^2）[(64k^2-24-48k^2)/(1+2k^2)^2]
=[2/(1+2k^2)]√（1+k^2）(4k^2-6)
S△OAB=（1/2）[2/√（1+k^2）]* [2/(1+2k^2)]√（1+k^2）(4k^2-6)
=2[√ (4k^2-6)]/（1+2k^2）,
Take the derivative of K and make it 0
8k（-2k^2+7）/(1+2k^2)^2=0,
2k^2=7,
k=±√14/2,
When the linear equation is y = (± √ 14 / 2) x + 2, the triangle OAB area is the largest
((1+sinθ-cosθ)/(1+sinθ-cosθ))+ cot(θ/2)
(1) Let t = Tan (θ / 2) from the universal formula sin θ = (2t) / (T ^ 2 + 1) cos θ = (T ^ 2-1) / (T ^ 2 + 1) original formula = (T ^ 2 + 1 + 2t-t ^ 2 + 1) / (T ^ 2 + 1 + 2T + T ^ 2-1) = (2t + 2) / (2t ^ 2 + 2t) = 1 / T = cot (θ / 2) (2) let t = Tan (3a / 2) Tan 3A = (2t) / (1-T ^ 2) original formula = t * (2t) / (1-T ^ 2) + 1 = (
As shown in figure 23-3-4, in the rectangular coordinate system, the quadrilateral oabc is a rectangle, and the coordinates of point C are (3,6). If point P moves from point O along OA to point a at a speed of 1cm / s,
Point Q moves at a speed of 2cm / s along AC from point A. if P and Q start from O and a at the same time, we ask: (1) after how long, the area of triangular PAQ is 2cm ^ 2? (2) can the area of triangular PAQ reach 3cm ^ 2?
Let me start with the first question
Let P and Q move by x seconds
∵C（3,6）
∴AO=3,AC=6
AP=3-x,OP=x,AQ=2x,CQ=6-2x
Let s triangle PAQ = 2,
∴1/2*AP*AQ=2
The solution is: X1 = 1, X2 = 2
That is, when X1 = 1 or x2 = 2
S △ PAQ = 2cm & # 178;
What about the picture? At least give the coordinates of A!
There is no picture
I also want this answer
I don't know either
Y = x, square LNX, find monotonicity according to derivative
y'=x(1+2Lnx)
y=x^2*lnx
The definition domain of the equation is:
{x|x>0}
y'=2x*lnx+x^2*1/x
=2x*lnx+x
Let y '>'0
2xlnx+x>0
x(2lnx+1)>0
X>0
therefore
2lnx+1>0
lnx>-1/2
x>e^(-1/2)
So when x belongs to (e ^ (- 1 / 2), positive infinity), the function is monotonically increasing
When x belongs to (0, e ^ (- 1 / 2)), the function is monotonically decreasing
If Tan α + cot α = 2, then Tan & sup2; α + cot & sup2; α = sin α * cos α=
tanα+cotα=2
Square on both sides:
tan²α+cot²α+2=4
tan²α+cot²α=2
tanα+cotα=2
sinα/cosα+cosα/sinα=2
1/sinα*cosα=2
sinα*cosα=1/2
tan²α+cot²α=(tanα+cotα)²-2tanαcotα
=2²-2=2；
Tan α + cot α = 2, then sin α / cos α + cos α / sin α = 2
The results show that sincos, α = 1 / α.
From the square of Tan α + cot α = 2, Tan & sup2; α + cot & sup2; α + 2tan α * cot α = 4, that is, Tan & sup2; α + cot & sup2; α = 2;
Cut the chord to tan α + cot α = 1 / sin α * cos α = 2
So sin α * cos α = 1 / 2
That is, Tan & sup2; α + cot & sup2; α = 2, sin α * cos α = 1 / 2.
As shown in the figure, in the plane rectangular coordinate system, O is the origin, the quadrilateral oabc is a rectangle, a (10,0), C (0,3), point D is the midpoint of OA, and point P moves on the side of BC. When △ ODP is an isosceles triangle with waist length of 5, the coordinate of point P is______ .
From the meaning of the question: od = 5 ∵ △ ODP is an isosceles triangle with waist length of 5 ∵ OP = 5 or PD = 5, make od vertical line through P, intersect with OD at point Q ∵ PQ = OC = 3 ∵ if OP = 5, then the right side OQ of right angle ∵ OPQ = 4, then the coordinates of point P are (4,3); if PD = 5, then QD = 4, OQ = 1, then the coordinates of point P are (1,3); if
If the line y = kx-1 is tangent to the curve y = LNX, then k = ()
A. 0B. -1C. 1D. ±1
∵ y = LNX, ∵ y '= 1X, let the tangent point be (m, LNM), the slope of tangent is 1m, so the tangent equation of curve at point (m, LNM) is y-lnm = 1m × (x-m). It passes (0, - 1), ∵ 1-lnm = - 1, ∵ M = 1, ∵ k = 1, so C is selected
Simplification of [cot (θ + 4 π) cos (θ + π) Sin & sup2; (θ + 3 π)] / [Tan (π + θ) cos & sup2; (- π - θ)]
The original formula = (- cot θ cos θ Sin & # 178; θ) / (Tan θ cos & # 178; θ)
=（-cotθsin²θ）/（tanθcosθ）
=（-cotθsinθtanθ）/tanθ
=-cotθsinθ
=-cosθ
[cot(θ+4π)cos(θ+π)sin²(θ+3π)]/[tan(π+θ)cos²(-π-θ)] =[cot(θ+4π)（-cosθ）sin²θ]/[tanθcos²θ]
=-cot(θ+4π）=（tanθ-1）/（1+tanθ）
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