Find the vertex trajectory equation under the ellipse with point a (1,2), eccentricity 12 and X-axis as the guide line

Find the vertex trajectory equation under the ellipse with point a (1,2), eccentricity 12 and X-axis as the guide line

Let f (x0, Y0) be the focus below the ellipse, the vertex below the ellipse be p (x, y) defined by | AF | 2 = 12, | AF | = 1, that is, the trajectory equation of point F is (x0-1) 2 + (y0-2) 2 = 1, and x0 = x, Y0 = 32y, the P trajectory equation of point F is (x − 1) 2 + (32y − 2) 2 = 1

A triangle and a parallelogram have the same base and area. The parallelogram is 6 meters high and the triangle is 3 meters high

Let the high h-base of a triangle be X
Because the areas are equal
xh/2=6x
h=12
So it's 12 meters high

Given two points m (- 1,0), n (1,0), if there is a point P satisfying PM · PN = 0 on the line 3x-4y + M = 0, then the value range of real number m is ()
A. （-∞，-5]∪[5，+∞）B. （-∞，-25]∪[25，+∞）C. [-25，25]D. [-5，5]

∵ two points m (- 1,0), n (1,0) if there is a point P on the line 3x-4y + M = 0 satisfying PM · PN = 0 ∵ this problem is transformed into the range of m when the line 3x-4y + M = 0 intersects the circle x2 + y2 = 1, that is, the distance from the origin (0,0) to the line 3x-4y + M = 0 is less than or equal to the radius, that is, | m | 32 + 42 ≤ & nbsp; 1, the solution is: - 5 ≤ m ≤ 5, so choose D

The width of the rectangle is 3A + 2B, and the length is more than the width a-b. how to find the perimeter of the rectangle?

Is there any result

The chord Ba is the chord of circle O, P is the point of the extension line on the chord Ba, PA = Pb = 2, Po = 4, find the radius of circle o
Wrong condition, PA = Ba = 2

O as OC vertical ab
Then CA = CB = 1 / 2Ab = 1
PC=2+1=3
OCP is a right triangle
OP is hypotenuse = 4
So OC ^ 2 = 4 ^ 2-3 ^ 2 = 7
With OCA, it's a right triangle
OA is a hypotenuse and a radius
So OA ^ 2 = OC ^ 2 + AC ^ 2 = 7 + 1 = 8
So radius = 2, root 2

As shown in the figure, in △ MNP, h is high and is the intersection of MQ and NE, and QN = QM. Prove that △ PQM is all equal to △ hQN
It is required to write the proof process steps in order

∵∠ meh = ∠ nqh = 90 ° (vertical definition),
∠ MHE = ∠ nhq (equal to vertex angle),
The remaining angles of equal angles are equal
MQ = NQ (known)
∠ mqp = ∠ nqh = 90 ° (known)
∴△MPQ≌△NHQ

As shown in the figure, in trapezoidal ABCD, ad ‖ BC, ab = CD, two diagonal lines AC and BD are perpendicular to each other, and the length of median line EF is 10, then the area of trapezoidal ABCD is ()
A. 200B. 20C. 100D. 50

The length of median line EF of ∵ trapezoid ABCD is 10, ∵ AD + BC = 2ef = 20. If DM ∥ AC intersection BC extension line is made at point m and DN ⊥ BC is made at point n, then ad = cm, ∵ AC ⊥ BD, ∵ BDM are isosceles right triangle, ∵ dn = 12 (BC + cm) = EF = 10, and ∵ EF is the median line of trapezoid, ∵ AD + BC = 2ef = 20

Let a = {X / X & # 178; + PX + q = O}, B = {X / X & # 178; + (p-1) x + 5-q = 0}, if a ∩ B = {1}?

The intersection is 1
Then 1 is the root of the two equations
So 1 + P + q = 0
1+p-1+5-q=0
So p = - 3, q = 2
A is X & # 178; - 3x + 2 = 0
x=1,x=2
Q is X & # 178; - 4x + 3 = 0
x=1,x=3
So a ∪ B = {1,2,3}

Let the moving straight line l be perpendicular to the x-axis and intersect with the ellipse x square + 2Y square = 4 at two points a and B. P is the point on L satisfying PA vector multiplied by Pb vector = 1, and the equation of P is obtained

Let P (x, y) then - 2

Given X4 + X3 + x2 + X + 1 = 0, find the value of 1 + X + x2 +... + x2010

Answer: 1 + X + x2 +... + x2010 = 1 + X + x2 + X3 + X4 + X5 (1 + X + x2 + X3 + X + x4) + x10 (1 + X + x2 + X3 + x4)... + x2005 (1 + X + x2 + X3 + x4) + x2010 = x2010x4 + X3 + x2 + x2 + X + 1 = 0, multiply both sides by (x-1), get: (x4 + X3 + x2 + X + 1) (x-1) = x5-1 = 0, so X5 = 1, so x2010 = (x5) ^ 402 = 1 ^ 402 = 1