The angle between two asymptotes of hyperbola x ^ 2-y ^ 2 / A ^ 2 = 1 (a > √ 2) is π / 3, and the eccentricity of hyperbola is calculated Say 2 √ 3 / 3?

The angle between two asymptotes of hyperbola x ^ 2-y ^ 2 / A ^ 2 = 1 (a > √ 2) is π / 3, and the eccentricity of hyperbola is calculated Say 2 √ 3 / 3?

Upstairs is right
The angle is π / 3, that is, the angle between an asymptote and the y-axis is π / 6, and the slope of the asymptote is tan60
That is B: a = tan60
And because C ^ 2 = a ^ 2 + B ^ 2
The centrifugal ratio C / A is 2

The straight line y = half of X + 2 intersects the X axis at point a, the Y axis at point B, and the hyperbola y = half of x m intersects at point C, CD ⊥ X axis at d; s △ ACD = 9, find 1 hyperbola

A (- 4,0) B (0,2) let the coordinate of point C be [x, (1 / 2) x + 2], then the coordinate of point D is (x, 0) because s △ ACD = 9. Drawing the picture, we can see that s △ ACD = s △ AOB + s △ BOC + s △ ODC = 1 / 2 | Ao |. | ob | + 1 / 2 | Bo |. | OD | + 1 / 2 | OD |. | CD | gets 1 / 2 * 4 * 2 + 1 / 2