It is known that hyperbola m passes through point P (4, √ 6 / 2), and its asymptotic equation is x ± 2Y = 0

It is known that hyperbola m passes through point P (4, √ 6 / 2), and its asymptotic equation is x ± 2Y = 0

Application of Hyperbolic Equations
Let the hyperbolic equation be
Square of X - square of 4Y = k
Substituting the coordinates of the points,
16-4·3/3=k
The solution is k = 10
So the hyperbolic equation is
Square of X - square of 4Y = 10

It is known that an asymptotic equation of hyperbola is x + 2Y = 0, and the standard equation of hyperbola is obtained by (2.2)

It is known that an asymptotic equation of hyperbola is x + 2Y = 0
Let the standard equation of hyperbola be (x + 2Y) (x-2y) = C
Substitute (2,2) to get C = - 12
So the standard equation of hyperbola is y * y / 3-x * x / 12 = 1

Given that the hyperbola passes through the point P (√ 5,1 / 2), the asymptotic equation is x ± 2Y = 0, and the focus is on the ± axis, the standard equation of the hyperbola is obtained

The hyperbolic asymptote is x ± 2Y = 0, let the hyperbolic equation be: X & # 178; - 4Y & # 178; = M
Substituting point P, we get m = 4
Namely:
x²－4y²=4
The standard equation is: X & # 178 / 4-y & # 178; = 1

If the asymptote equation of hyperbola x square / 4-y square / b square = 1 is 3x-2y = 0, what is the value of B?
Happy new year to you all

The asymptote of hyperbola x square / 4 - y square / b square = 1 is
x/2±y/b=0
It is known that the asymptote is x / 2-y / 3 = 0
The comparison coefficient is b = ± 3