## Algebra and Trigonometry 10th Edition

$\dfrac{x^2}{1}-\dfrac{y^2}{4}=1$
The standard form of the equation of the hyperbola with a horizontal transverse axis can be expressed as: $\dfrac{(x-h)^2}{a^2}-\dfrac{(y-k)^2}{b^2}=1$ The vertices and foci have the form $(\pm a, 0)$ and $(\pm c,0)$. The standard form of the equation of the hyperbola with a vertical transverse axis can be expressed as: $\dfrac{(y-k)^2}{a^2}-\dfrac{(x-h)^2}{b^2}=1$ The vertices and foci have the form $(0, \pm, a)$ and $(0, \pm c)$. We have: $a=1$ $y=\dfrac{bx}{a} \implies 2x=\dfrac{bx}{1}$ or, $b=2$ $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$ or, $\dfrac{x^2}{1}-\dfrac{y^2}{4}=1$