Given that the center of the ellipse is at the origin, e = √ 3 / 2, and one of its focuses coincides with the focus of the parabola x ^ 2 = - 4 √ 3Y, then the equation of the ellipse is Is x ^ 2 / 3 + y ^ 2 = 1 right

Given that the center of the ellipse is at the origin, e = √ 3 / 2, and one of its focuses coincides with the focus of the parabola x ^ 2 = - 4 √ 3Y, then the equation of the ellipse is Is x ^ 2 / 3 + y ^ 2 = 1 right

The focus is on the y-axis, at this time e = C / B, B = 2, C = 3 ^ (1 / 2), a = 1, should be x ^ 2 + y ^ 2 / 4 = 1
It is known that the ellipse e passes through the point a (2,3), the center is at the origin, the focus is on the x-axis, the eccentricity e = 1 / 2, (a). The equation for solving the ellipse (has been calculated, with the emphasis on the following)
(b) If F1 and F2 are the left and right focal points of the ellipse, try to find the equation of the straight line where the bisector of the angle f1af2 is located
There should be a detailed process, please ~ ~ ~ good bonus points
（1）∵e=c/a=1/2∴a=2c,b²=3c²
Ψ ellipse X & # / 4C & # / 178; + Y / 3C & # / 178; = 1  a  4 / 4C & # / 178; + 9 / 3C & # / 178; = 1  C = 2  ellipse X & # / 178; + 16 + Y & # / 178; = 1
(2) Suppose F1 (- 2,0), F2 (2,0). Then af2 ⊥ F1F2. F1F2 = 4, af2 = 3. According to Pythagorean theorem, AF1 = 5
Let the angle bisector of ∠ f1af2 intersect F1F2 at M. then according to the angle bisector theorem, f1m / MF2 = F1A / af2
That is, m (0.5,0)
Let am: y = KX + B. then
2k+b=3,0.5k+b=0∴k=2,b=-1
The obtained line: y = 2x-1
Given that the center of the ellipse is at the origin and passes through the point P (3,2), the focus is on the coordinate axis, and the length of the major axis is three times that of the minor axis, the equation of the ellipse is solved
1. Focus on the x-axis, let the elliptic equation be x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1, the length of the major axis is three times the length of the minor axis: a = 3b, because of P (3,2), substituting 9 / A ^ 2 + 4 / B ^ 2 = 1, the solution is a ^ 2 = 45; B ^ 2 = 52, focus on the y-axis, let the elliptic equation be y ^ 2 / A ^ 2 + x ^ 2 / b ^ 2 = 1, a = 3B 4 / A ^ 2 + 9 / b ^ 2 = 1, the solution is a ^ 2
1. The focus is on the X axis
Let: the elliptic equation be: x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1
The length of major axis is three times of that of minor axis: {2A = 3 * 2B
Because constant over P (3,2): {9 / A ^ 2 + 4 / b ^ 2 = 1
The solution is as follows
{a=
{b=
2. The focus is on the y-axis
Let the elliptic equation be y ^ 2 / A ^ 2 + x ^ 2 / b ^ 2 = 1
In the same way:
{2A=3*2B
{9 / A ^ 2 + 4 / b ^ 2 =... Expand
1. The focus is on the X axis
Let: the elliptic equation be: x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1
The length of major axis is three times of that of minor axis: {2A = 3 * 2B
Because constant over P (3,2): {9 / A ^ 2 + 4 / b ^ 2 = 1
The solution is as follows
{a=
{b=
2. The focus is on the y-axis
Let the elliptic equation be y ^ 2 / A ^ 2 + x ^ 2 / b ^ 2 = 1
In the same way:
{2A=3*2B
{9/a^2+4/b^2=1
The solution is as follows
{a=
{b=
So the equation of ellipse is:
Do it yourself! Training speed! Put it away
The equation of ellipse is the standard equation
(1) When the focus is on the x-axis, let the elliptic equation be x2a2+
y2b2=1(a＞b＞0)
Then 2A = 3 × 2b9a2+
4B2 = 1 # 8203;, and A2 = 45b2 = 5 # 8203;
The equation of ellipse is x245+
y25=1；
(2) When the focus is on the x-axis, let the elliptic equation be x2b2+
y2a2=1(a＞b＞0)
... unfold
The equation of ellipse is the standard equation
(1) When the focus is on the x-axis, let the elliptic equation be x2a2+
y2b2=1(a＞b＞0)
Then 2A = 3 × 2b9a2+
4B2 = 1 # 8203;, and A2 = 45b2 = 5 # 8203;
The equation of ellipse is x245+
y25=1；
(2) When the focus is on the x-axis, let the elliptic equation be x2b2+
y2a2=1(a＞b＞0)
Then 2A = 3 × 2b9b2+
4a2 = 1 &;, A2 = 85b2=
859​
The elliptic equation is 9x285+
y285=1；
In conclusion, the elliptic equation is x245+
Y25 = 1 or 9x285+
Y285 = 1
As shown in the figure, in the plane rectangular coordinates, the quadrilateral oabc is isosceles trapezoid, CB / / OA, OC = AB = 4, BC = 6, angle COA = 45 degrees, the moving point P starts from point O, moves on the edge of the trapezoid oabc, the path is o to a to B to C, and stops when it reaches point C. make a straight line CP
(1) Finding the area of trapezoid oabc
(2) When the straight line CP divides the area of trapezoidal oabc into two equal parts, the analytical expression of straight line CP is obtained
(3) When the triangle OCP is an isosceles triangle, write the coordinates of point P
The answer to the last question is (4,0), (4 radical 2,0), (2 radical 2,0) (4 + 2 radical 2,2 radical 2) I want process, if not, thank you, but (1), (2) must process
Thank you very much ~ 0 (T-T) 0
1. Because: angle COA = 45 degrees, so: CD = OC / √ 2 = 22, so: trapezoid area = (BC + OA) * CD / 2 = (6 + 6 + 2 * 2 * √ 2) * 2 * √ 2 / 2 = 12 √ 2 + 82, straight line CP divides the area of trapezoid oabc into two equal parts, s △ OPC = 1 / 2 * OP * CD, so: P point: 6 + 2 √ 2, straight line CP passes through two points: 2
(1) Because the angle COA = 45 degrees, OC = AB = 4, the height of the trapezoid is 2 times root number 2, and the bottom OA is 6 + 4 times root number 2, so the trapezoid area is 12 times root number 2 + 8.
(2) Let CP and X intersect at P. Then the area of triangle OCP is 6 times root 2 + 4, the height is 2 times root 2, so the bottom OP is 6 + 2 times root 2. The straight line CP passes through two points c and P, C (2 times root 2, 2 times root 2), P (6 + 2 times root 2, 0); so the analytical formula of the straight line CP is: y = negative 3 times root 2x + 2 times root 2 and 3 minutes... Expansion
(1) Because the angle COA = 45 degrees, OC = AB = 4, the height of the trapezoid is 2 times root number 2, and the bottom OA is 6 + 4 times root number 2, so the trapezoid area is 12 times root number 2 + 8.
(2) Let CP and X intersect at P. Then the area of triangle OCP is 6 times root 2 + 4, the height is 2 times root 2, so the bottom OP is 6 + 2 times root 2. The straight line CP passes through two points c and P, C (2 times root 2, 2 times root 2), P (6 + 2 times root 2, 0); so the analytical formula of the straight line CP is: y = the sum of minus 3 times root 2x + 2 times root 2 and 4 / 3. Put it away
If the function y = LNX is tangent to the line y = KX, then K=______ .
Let the tangent point be (x0, Y0), then ∵ y ′ = (LNX) ′ = LX, the tangent slope k = 1x0, and the point (x0, lnx0) is on the straight line, which is substituted into the equation to get lnx0 = 1x0 · x0 = 1, ∵ x0 = e, ∵ k = 1x0 = 1E
Simplification: radical (1-tan θ) cos & sup2; θ + (1 + cot θ) Sin & sup2; θ
Simplification
Radical (1-tan θ) cos & sup2; θ + (1 + cot θ) Sin & sup2; θ
(1-tanθ)cos²θ+（1+cotθ)sin²θ =cos²θ-sinθcosθ+sin²θ+sinθcosθ=1
One
The root of the trapezoid is oacb = 3, OC = 5, y = 6, and the axis of the trapezoid is parallel to oacb
Find the coordinates of point B in the rectangular plane coordinate system
Make BP ⊥ X axis and point P through point B
∵OA=6,CB=3
∴AP=3
∵ BP ⊥ X axis
∴∠APB=90°
∵ in RT △ ABP, ∠ APB = 90 °
AP = 3, ab = 3, radical 5
Ψ BP = (3 radical 5) & sup2; - (3) & sup2; = 6
∴B（3,6）
）△OAD∽△CDB. △ADB∽△ECB
（2）①（1，-4a）
②∵△OAD∽△CDB
Qi
∵ ax2-2ax-3a = 0, a (3, 0) can be obtained
OC = - 4A, OD = - 3a, CD = - A, CB = 1,
So the analytical formula of parabola is
③ Existence,
Let P (x, - x2 + 2x + 3)
∵△ pan is similar to △ oad, and △ oad is an isosceles triangle ∵ p... expansion
）△OAD∽△CDB. △ADB∽△ECB
（2）①（1，-4a）
②∵△OAD∽△CDB
Qi
∵ ax2-2ax-3a = 0, a (3, 0) can be obtained
OC = - 4A, OD = - 3a, CD = - A, CB = 1,
So the analytical formula of parabola is
③ Existence,
Let P (x, - x2 + 2x + 3)
∵ △ pan is similar to △ oad, and △ oad is isosceles triangle ∵ PN = an
When x0 (x > 3), x-3 = - (- x2 + 2x + 3), X1 = 0, X2 = 3
The qualified point P is (- 2, - 5) folded
If solution (1) is taken as BH ⊥ X axis at point h, then the quadrilateral ohbc is a rectangle,
Ψ Oh = CB = 3, (1 point)
∴AH=OA-OH=6-3=3
In RT △ ABH, BH = = 6 (2 points)
The coordinate of point B is (3.6) (3 points)
(2) Let eg ⊥ X be at point G, then eg ∥ BH
■ △ OEG ∽ obh (4 points)
∵ OE = 2EB
∴ ，∴ = ，
∴OG=2，EG=4
If the line y = KX is the tangent of y = LNX, then the value of K is
Otherwise I don't understand
Bonus points!
y=lnx
y'=1/x
Let (x0, Y0) be the tangent point
Then: k = 1 / x0
y0=kx0=1,x0=1/k
And: Y0 = lnx0
1=ln(1/k)
1/k=e
k=1/e
1 / E, let the tangent point (a, LNA), k = 1 / A, then LNA = a * 1 / A, so a = e, k = 1 / E
Given cot α = 1 / 3, find the value of Cos2 α
According to the known COTA = cosa / Sina = 1 / 3
Then Sina = 3cosa, Sin & # 178; a = 9cos & # 178; a
And Sin & # 178; a + cos & # 178; a = 1
Then cos & # 178; a = 1 / 10, Sin & # 178; a = 9 / 10
So cos2a = cos & # 178; a-SiN & # 178; a = - 4 / 5
In the plane rectangular coordinate system, the quadrilateral oabc is a trapezoid OA parallel CB, the coordinate of point a is 〈 6,0 〉, the coordinate of point B is {3,4], point C is on the y-axis, the moving point m moves on the OA side, starting from point O to point a; the moving point n moves on the AB side, starting from point a to point B, two moving points, starting at the same time, the speed is 1 unit length per second, when one of them reaches the end, the other stops immediately, Let the motion time of two points be T seconds
Q: if AAC is connected, is there such a value of t that Mn and AC are perpendicular to each other?
The line AB is y = (- 4 / 3) x + 8
The slope of AC is - 2 / 3. To make Mn perpendicular to AC, the slope of Mn is 3 / 2. M (T, 0) can get the linear equation of Mn as y = (3 / 2) x - (3 / 2) T. find out the intersection of line Mn and line AB, and then replace it with line AB to get t. test whether T meets the condition, and t belongs to [0,5] (m, n should be on OA, ab)
Let the line Mn be perpendicular to the line AC. The product of slopes is negative one.
A, there is such a t that Mn and AC are perpendicular to each other.
It's better to draw a picture on grass paper first
1. According to the meaning of the title, the coordinates of point C are (0,4)
2. AB = 5, Ao = 6, the velocity of VM and VN is the same, so m can go to (5,0) and N to B at most.
Mn changed from Ao to (5,0) B.
There is.
If Mn and AC are perpendicular to each other, the slope product is - 1
A(6,0)，C(0,4)，M(t,0)，N(6-3/5t,4/5t)
The slope of AC is - 2 / 3, and the slope of Mn is 2T / (15-4t). It can be seen that 2T / (15-4t) = 3 / 2, and 16t = 45,
t=45/16 =2.8125