If a is the inclination angle of line L and Sina + cosa = 0.2, what is the slope of line l?

If a is the inclination angle of line L and Sina + cosa = 0.2, what is the slope of line l?

So: Sina ^ 2 + (0.2-sina) ^ 2 = 1, Sina = o.8 or - 0.6, slope = taga = Sina / cosa = 0.8 / (0.2-0.8) = - 1.33 or = taga = Sina / cosa = - 0.6 / (0.2 - (- 0.6)) = - 1.5

Given that the inclination angle of the line L is a, Sina + cosa = 1 / 5, then the slope of L is?

The sum of the squares of sin α + cos α = 1 / 5 gives 1 + 2 sin α cos α = 1 / 25
That is sin2 α = - 24 / 25

It is known that the two focal points of the ellipse x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 (a > b > 0) are F1 (- C, 0), F2 (C, 0) (c > 0), the straight line passing through point E (a ^ 2 / C, 0) intersects the ellipse at points a and B, and F1A / / f2b, | F1A | = 2 | f2b |
1. Find the eccentricity of the ellipse. 2. Find the slope of the straight line ab. 3. Let point C and point a be symmetrical about the origin of the coordinate, and there is a point H (m, n) (m ≠ 0) on the straight line f2b on the circumscribed circle of △ af1c, find the value of N / m

(1) Absolutely wrong, it should be | f2b | = 2 | F1A |. You can draw a picture to know that there is something wrong with the title
F1A / / f2b, then △ eaf1 ∽ ebf2
|EF1|/|EF2|=|F1A|/|F2B|=1/2
|EF1|=a²/c-c,|EF2|=a²/c+c
We get a & # 178; = 3C & # 178;
Then E = C / a = √ 3 / 3
(2) Let a (x1, Y1) B (X2, Y2) pass through the vertical lines AC and BD of a and B as the right guide line respectively, and B as the vertical line BH of the left guide line
If △ eaf1 ∽ ebf2, then y2 = 2y1
Then, | BD | = 2 | AC |, according to the second definition of ellipse
The distance from B to the left guide line | BH | = | BF2 | / E, and the distance from a to the right guide line | AC | = | AF1 | / E
|F2B|=2|F1A|,|BH|=2|AC|
So | BH | = | BD |, if the distance from point B to the left and right guide lines is equal, then B must be on the y-axis,
Then the coordinates of B are (0, b)
Straight line AB passes through point E (A & # 178 / C, 0), point B (0, b), e = √ 3 / 3, AB slope k = - BC / A & # 178; = - √ 2 / 3
The equation is y = - √ 2 / 3 + B

To solve the elliptic equation: the two focuses are F1 (- 2,0), F2 (2,0), and through (5 / 2, - 3 / 2)