The hyperbolic equation with the crossing point 1,1 and the asymptote equation of y 2x is ???

1) When the focus is on the x-axis.

Let the equation of the hyperbola be x a -y b = 1, then its asymptote is y= bx a, so there is b a = 2, b = 2a and hyperbola (1, 1) so 1 a -1 b = 1 conjunction, the solution is 1 a = 4 3 1 b = 1 3 so the equation of the hyperbola is 4x 3-y 3 = 12) when the focus is on the y axis.

y²/a²-x²/b²=1

Then its asymptote is y=ax b, so there is a b=2, a=2b and hyperbolic (1,1), so 1 a -1 b =1 is combined, and the solution is 1 b 0, which does not conform and is rounded.

So the equation for hyperbola is 4x 3-y 3=1

From the meaning, b = 2a or a = 2b, the solution is a = 4 3 or 16 3, b = 4 3 or 16 3, so the equation is x 4 3-y 16 3 = 1, x 16 3-y 4 3 = 1, o( o, hope it helps you, hope.

Solution: The asymptote equation is y= 2x, let the hyperbolic equation be 4x, -y=k, the passing point(1,1), 4-1=k, k=3, so the equation is 4x, -y=3, i.e., x(3, 4)-y, 3=1, 45, 1

According to the title, the equation for an asymptote of the hyperbola is y=x, and the circle is known.

Let the hyperbolic orange elimination equation be x2

y2 (0), hyperbolic point p(2,4), 22 ( cherry blossom 0), i.e. =-12

The hyperbolic equation is y2

x212, so the answer is: y2x2

According to the asymptotic equation of the hyperbola is 2x 3y=0, let the equation of the hyperbola be (2x+3y)(2x-3y)= 0), that is, 4x2-9y2= (0), the point (1,2) on the hyperbola, 4 12-9 22= , the solution =-32 From this, the equation of the hyperbola is 4x2-9y2=-32, which is simplified.

The equation for an asymptote of the hyperbola is y=x, and the hyperbolic equation can be x2 -y2=k,(k≠0).

The points (2,4) are taken with Tanzhou on the hyperbolic stupid line, and the hyperbolic equation is substituted into the hyperbolic equation, and the standard hyperbolic equation of 4-16=kk=12 is y 2 -x 2 =12, so the answer is the letter y 2 -x 2 =12

Solution: The asymptote equation is y= 2x

Let the hyperbolic equation be 4x -y =k

Crossing points (1, 1).

4-1=kk=3

So Eq. 4x -y =3

i.e. x (3 4)-y 3=1

1。If the equation of an asymptote line of the hyperbola is y 1 2x, and it passes through the point p(3,1 2), then one of its asymptotic equations is y 1 2x,b a=1 2,a=2b, and the hyperbolic equation is: x 2 4

y=±a/bx=±1/2

The square of x, the square of b-the square of y, the square of a=1, the square of 16 4a-12 the square of 4a=1

The square of a is 1

Let the hyperbolic equation be 9x -y = k

Substituting (2,-1) yields 36-1=k

The hyperbolic equation is 9x -y = 35

This process is not easy to come out, let me tell you the method. Since you don't know which axis it's focused on, it needs to be discussed in categories. First of all, it is assumed that on the x-axis, we can get that b is equal to 3 than a, and then bring in the coordinates of the point p, and we can get a relationship between a and b, and the equation can be solved.

Then set the focus on the y-axis, the same method.