What is the equation of this hyperbola? What is the equation of the hyperbola that satisfies these conditions:
Asymptotes $y=2x$ and $y=-2x$, centre $(0,0)$, and the point $(1,1)$ lies on the curve.
This isn't a homework question; I study maths for the fun of it. The question is taken from the book "Delta Mathematics" by David Barton and Anna Cox, 2013 edition, page 22 question 11 b.
The answer in the back of the book is $$\frac{4x^2}{3} - \frac{y^2}{3} = 1$$
But how do I work out the answer from the given information?  
I started by sketching a graph to see where the point $(1,1)$ falls.

I don't know how to continue at this point.
 A: The asymptotes of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ are $y = \pm \frac{bx}{a}$. 
Since $(1, 1)$ lies on your curve then $\frac{1}{a^2} - \frac{1}{b^2} = 1 \iff b^2 -a^2 = a^2b^2$. 
But you know that $\frac{b}{a} = 2\iff b = 2a$. Plug this into the above equation to get ($a\neq 0$) $$3a^2 = 4a^4 \Rightarrow a^2 = \frac{3}{4}$$ and $b^2 = 3$. 
So the hyperbola is $$\frac{x^2}{\frac{3}{4}} - \frac{y^2}{3} = 1$$
A: An equation of the hyperbola  centred at $(0,0)$, with the coordinate axes as axes of symmetry is: 
$$\textsf{product of the equations of the asymptotes = constant}.$$
Here, you have
$\;y^2-4x^2=1^2-4\cdot1^2=-3$, or, in normalised form
$$\frac43x^2-\frac13y^2=1.$$
A: The slope of the asymptotes is $\pm \frac{b}{a}=\pm 2 \implies b=2a$
All hyperbola with an equation like $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ are centered on the origin. The hyperbola does contain the point $(1,1)$ too, so
$\frac{1}{a^2}-\frac{1}{b^2}=1$
$\frac{1}{a^2}-\frac{1}{4a^2}=1$
$\frac{3}{4a^2}=1$
$a^2=\frac{3}{4}$ and 
$b^2=4a^2=4\frac{3}{4}=3$ 
