The abscissa of a point P on the ellipse is 3, and the distances from it to the two focal points are 6.5 and 3.5, respectively Such as the title

The abscissa of a point P on the ellipse is 3, and the distances from it to the two focal points are 6.5 and 3.5, respectively Such as the title

6.5+3.5=2*a
A=5
I don't know if you taught this formula
The distance between a point on the ellipse and two focal points is
A + X * C / A and A-X * C / A
So C = 5 / 2
b2=75/4
You can get an ellipse
The square of sina + Sina = 1, the square of 3cosa + the fourth power of COS - the value of 2sina + 1
Because Sina + Sina ^ 2 = 1, so Sina = 1-sina ^ 2
3cosa^2+cosa^4-2sina+1
=3（1-sina^2）+(1-sina^2)^2-2sina+1
=3sina+sina^2-2sina+1
=sina^2+sina+1
=If you don't understand, please ask,
The sum of the distances from the point a (1,3 / 2) on the ellipse C: x2 / A2 + Y2 / B2 = 1 to the two focal points is 4 (1)
(2) Let K be the moving point on the ellipse in (1) and F1 be the left focal point. The trajectory equation of the midpoint of the line f1k is obtained
How to do this kind of question? I can't do it half way
The first one asked me what I did
Since the first one can do it, let's not talk about the process. Write the equation as X & # 178 / 4 + Y & # 178 / 3 = 1. Second question: let's write the coordinates of F1 (- 2,0), then the coordinates of K point is (2x + 2,2y). Since K is on the ellipse, it satisfies the elliptic equation, that is, (2x + 2) &# 178 / 4 + (2Y) &# 178 / 3 = 1
If sin α: cos α / 2 = 4:5, then Tan α / 4=
Replace sin a with sin half a
And then we eliminate COS and get sin, and then we get sin quarter a from this
Then we calculate cos tan
It is known that the left and right focal points of the ellipse x ^ 2 / 4 + y ^ 2 / 3 = 1 are F1F2 respectively. A straight line L passes through F1 and intersects the ellipse at two points a and B. if the inclination angle of L is π / 4,
Finding the area of △ Abf
A straight line L passes through the F1, and the inclination angle is π / 4, y = x + X + 1x = y-13 (Y-1) \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\myface
The focus F1, F2 coordinates are easy to get (1,0) (- 1,0)
No matter which focus you pass, the area is the same
Let F 1 (1,0), then the equation of L is y = X-1
Let the coordinates of intersection be (x1, Y1) (X2, Y2)
In the elliptic equation
（y+1）²/4+y²/3=1
Its area = | F1F2 | (| Y1 | + | Y2 |) / 2, | F1F2 | = 2
It is obvious that | Y1 | + | Y2 | = | y1-y2 |
The focus F1, F2 coordinates are easy to get (1,0) (- 1,0)
No matter which focus you pass, the area is the same
Let F 1 (1,0), then the equation of L is y = X-1
Let the coordinates of intersection be (x1, Y1) (X2, Y2)
In the elliptic equation
（y+1）²/4+y²/3=1
Its area = | F1F2 | (| Y1 | + | Y2 |) / 2, | F1F2 | = 2
It's obvious that | Y1 | + | Y2 | = | y1-y2|
Y1, Y2 are the quadratic equations of one variable (y + 1) &# 178 / 4 + Y & # 178 / 3 = 1
According to Weida's theorem
y1+y2=-6/7
y1*y2=-9/7
So it's easy to get
|y1-y2|=√（（y1+y2）²-4y1y2）=12√2/7
So area = 12 √ 2 / 7
Abf1 or abf2?
Tan (π + α) = - 1 / 2 for sin (α - 7 π) · cos (α + 5 π)
tan(π+α)=tanα=-1/2 sin(α-7π)·cos(α+5π) =-sinα*（-cosα） =sinα*cosα/[(sinα)^2+(cosα)^2] =tanα/[(tanα)^2+1] =-1/2/[(-1/2)^2+1] =-2/5
If we simplify the formula, we get Tana = - 1 / 2, because (Sina) ^ 2 + (COSA) ^ 2 = 1, we get the root 5 of SIA = 5, cosa = - 2 times of 5, or Sina = - 5 times of 5, cosa = 2 times of 5, the original formula is Sina times (- COSA) = - 2 of 5
The left and right focal points of the ellipse x2a2 + y2b2 = 1 (a > b > 0) are F1 and F2 respectively. The inclination angle passing through the focal point F1 is 30 ° and the straight line intersects the ellipse at two points a and B. the chord length | ab | = 8. If the area of the inscribed circle of the triangle abf2 is π, the eccentricity of the ellipse is ()
A. 22B. 36C. 12D. 33
The equation of the straight line AB is y = 33 (x + C), that is, x-3y + C = 0, the distance from F2 to the straight line AB is d = 2c2 = C, the area of the inscribed circle of the triangle abf2 is π, then the radius is 1, and 12 × 8 × C = 12 × 4A × 1, and х e = CA = 12 can be obtained from the equal area
If sin α cos α = 2 / 5, then Tan α?
sinαsinα+cosαcosα=1
sinαcosα=2/5
You can get it by going up and down
tanα+1/tanα=5/2
therefore
Tan α = 2 or 1 / 2
Through the left focus F1 of the ellipse x ^ 2 / 2 + y ^ 2 = 1, make a straight line L with an inclination angle of 60 degrees. The straight line L and the ellipse intersect at two points a and B. find the length of AB and the direction of Da Xian!
F1F2 is the diameter of the circle, pf1f2 is a right triangle, Pf1 + PF2 = 2A, Pf1 * PF2 / 2 = 26. F1F2 = 2C
(2c)^2=PF1^2+PF2^2=(PF1+PF2)^2-2PF1*PF2=121
c=11/2
a=15/2
b=√26
a+b+c=13+√26
I've written down your serial number. I'll make it for you after adoption
Tan @ = 3, and @ is the angle of the third quadrant, find sin @ cos@
A is the angle of the third quadrant
sina/cosa=3
So Sina = 3cosa
From (Sina) ^ 2 + (COSA) ^ 2 = 1
sina=-3√10/10 ,cosa=-√10/10
Sin @ = 3 / √ 10 (root 10)
cos@= -1√10