[sharp!] hyperbola x ^ 2 / A ^ 2-y ^ 2-B ^ 2 = 1, if the inclination angle of an asymptote is 3 / π and the eccentricity is e, then the minimum value of (a ^ 2 + e) / b RT, the answer should be (2 √ 6) / 3, The problem is wrong, it should be x ^ 2 / A ^ 2-y ^ 2 / b ^ 2 = 1

[sharp!] hyperbola x ^ 2 / A ^ 2-y ^ 2-B ^ 2 = 1, if the inclination angle of an asymptote is 3 / π and the eccentricity is e, then the minimum value of (a ^ 2 + e) / b RT, the answer should be (2 √ 6) / 3, The problem is wrong, it should be x ^ 2 / A ^ 2-y ^ 2 / b ^ 2 = 1

The inclination angle of an asymptote is π / 3 = = > b / a = √ 3 = = > b = √ 3aC ^ 2 = a ^ 2 + B ^ 2 = = > C ^ 2 = 4A ^ 2 = = > C = 2A = = > e = 2 [(a ^ 2 + E) / b] ^ 2 = (a ^ 2 + 2) ^ 2 / (3a ^ 2) = (a ^ 4 + 4A ^ 2 + 4) / (3a ^ 2) = 1 / 3 (a ^ 2 + 4 / A ^ 2 + 4) ≥ 1 / 3 [2 √ (a ^ 2 * 4 / A ^ 2) + 4] = 1 / 3 (4 + 4) = 8 / 3, when a ^ 2 = 4 / A ^ 2, take the equal sign

The eccentricity of hyperbola with focus on x-axis is e = 2, and the inclination angle of an asymptote is 60 degrees

The inclination angle is 60 degrees
Then k = B / a = tan60 = √ 3
b²=3a²
c²=a²+b²=4a²
Then E = C / a = 2
So it's actually the same condition
So we can't find out
X & # 178 / / A & # 178; - Y & # 178 / / 3A & # 178; = 1, where a > 0

Find the hyperbolic equation which passes through point (3, - 2) and the inclination angle of an asymptote is π / 6

The slope is k = Tan π / 6 = √ 3 / 3
The asymptote is y = ± (B / a) X
So B / a = √ 3 / 3
a=√3b
a²=3b²
x²/a²-y²/b²=±1
Substituting
9/3b²-4/b²=±1
-1/b²=±1
So take a minus sign on the right, B & sup2; = 1
a²=3
So y & sup2; - X & sup2 / 3 = 1

Given that the angle between two asymptotes of hyperbola x2a2 − y2b2 = 1 (b > a > 0) is π 3, then the eccentricity of hyperbola is______ ．

As shown in the figure, the angle between the two asymptotes of M, n. ∵ hyperbola x2a2 − y2b2 = 1 is π 3, and B ∵ a ∵ a ∵ 0, ∵ BA ∵ 1, ∵ should be ∵ mon = π 3. And ∵ MOX ∵ π 4. ∵ MOX = π 2 − 12 × π 3 = π 3. ∵ Ba = Tan ∵ MOX = t