If the inclination angle of the line L is a and sin2a = 3 / 5, the slope of the line is

If the inclination angle of the line L is a and sin2a = 3 / 5, the slope of the line is

sin2a=3/5,
Obviously 0

1、 It is known that the coordinates of two points on the line L are the slope k of the line a (1,3) B (3,5)=_____ Inclination angle=_______ 4. If

(5-3)/(3-1)=1
Tan 45 ° = 1, so the tilt angle is 45 degrees

If the inclination angle of the line is a and Sina + cosa = - 1 / 5, then the slope of the line
The answer is - 4 / 3

Very simple, Sina square + cosa square = 1, Sina square + 2sinacosa + cosa = 1 / 25, then we can get sinacosa = - 12 / 25, so we can get two kinds of answers, Sina = 3 / 5, cosa = - 4 / 5, or Sina = - 4 / 5, cosa = 3 / 5, then Tana = - 3 / 4
Or Tana = - 4 / 3, that is, the slope of the line is - 3 / 4 or - 4 / 3

What is the slope of the line passing through two points a (- 3,5) B (0,2)?

Y = KX + B 5 = - 3K + B 2 = b 5 = - 3K + 2 - 3K = 3 K = - 1 the slope of the straight line passing through two points a (- 3,5) B (0,2) k = - 1

The focus of the ellipse is F1F2, the line passing through the point F1 intersects the ellipse, the shortest segment Mn cut by the ellipse is 32-5, the perimeter of the triangle mf2n is 20, and the eccentricity is calculated

According to the known conditions, Mn is perpendicular to the x-axis
MF1+MF2=2a
NF1+NF2=2a
Triangle mf2n perimeter = MF1 + MF2 + NF1 + NF2 = 4A = 20
a=5
MF1+MF2=2a
(MF1)^2+ (2c)^2=(MF2)^2
MF1=8/5
So C = √ 17
e=c/a=√17/5

It is known that the eccentricity of ellipse C: x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 (a > b > 0) e = √ 2 / 2, the left and right focus are f1.f2, the fixed point P (2, √ 3), and | F1F2|=

If P is on an ellipse, then 4 / A & sup2; + 3 / B & sup2; = 14b & sup2; + 3A & sup2; = A & sup2; B & sup2; (1) e = C / AE & sup2; = C & sup2; / A & sup2; = 1 / 2A & sup2; = 2C & sup2; a & sup2; = B & sup2; + C & sup2; 2C & sup2; = B & sup2; + C & sup2; B & sup2; = C & sup2; b = C is substituted into (1) 4C & sup2

Given that point P is the point on the ellipse x ^ 2 / 25 + y ^ 2 / 9 = 1, the area of triangle with point P and focus F1, F2 as vertex is equal to 8

According to the elliptic equation, if a ^ 2 = B ^ 2 + C ^ 2, then C = 4, f 1F 2 = 8, and the area of the triangle is 8, then the height of the triangle is 2, then p (x, y) | x | = 2, x = 2 or - 2, substitute x ^ 2 = 4 into the original equation to get y = plus minus three fifths root sign 21

Point P is an ellipse x ^ 2 / 25 + y ^ 2 / 9 = 1. The area of the triangle with point P and focus F1 and F2 as vertices is equal to 8. Calculate the coordinates of point P~

C ^ 2 = 25-9 = 16, C = 4,2c = 8,8 = 1 / 2 * 8 * ABS (YP), (ABS (YP)) = 2. Substituting the ellipse, we can get ABS (XP) = SQR (125) / 3. Therefore, the coordinates of point P are (± 5sqr (5) / 3, ± SQR (2))

The left and right focal points of the function ellipse x2 / 4 + Y2 / 2 = 1 are F1 and F2 respectively. The line L crosses F2 and the ellipse at two points AB, and O is the origin of the coordinate
Let a circle with diameter AB just pass through O, and find the equation of L

Let a (x1, Y1), B (X2, Y2). Then X1 * x2 + Y1 * y2 = 0; let the linear equation be y = KX + B, and the linear equation pass the right focus (√ 2,0). Then the linear equation can be reduced to y = KX - √ 2K

It is known that F1 and F2 are the two focal points of the ellipse x2 / 25 Y2 / 9 = 1. If the line passing through the point F1 intersects the ellipse at two points a and B, then the perimeter of the triangle abf2 is?

Two focus F1F2 distance = 2C = 8
AF1 + af2 = 2A = 10 (known by definition)
The perimeter of △ af1f2 = 8 + 10 = 18