Find the equation of a hyperbola with the focus at $(5,1)$ and asymptote at $y=1\pm 2x$
I know that the hyperbola would be a vertical one, since the asymptote is at $y$. The asymptote's formula is $$y = -(a/b)x + k+(a/b)h.$$ (The $-/+$ are interchangeable.)
But I can't seem to work out on what to do. Please help! Thanks! :-)
Let's see...
1) Draw the asymptotes and you see that they are symmetric around lines that are parallel to or coincident with the $x$ and $y$ axes. So we are sure that the hyperbola will have an equation with the form:
${(x-c)^2}/{a^2}-{(y-d)^2}/{b^2}=\pm 1$
Your task is then to find the parameters $a,b,c,d$ and whether the right side id $+1$ or $-1$.
2) Find the center as the intersection of the asymptotes as directed by @nftaussig. This gives you $c$ and $d$ as the coordinates of the center.
3) Compare the given focus with the center. The focus will be displaced horizontally or vertically from the center. Horizontal means the right side of the equation is $+1$, vertical means the right side is $-1$.
4) The distance from the center to either focus is $\sqrt{a^2+b^2}$. Note the sign difference from an ellipse where it's $\sqrt{a^2-b^2}$.
5) The asymptotes have slope $\pm(b/a)$. Use this and (4) to get $a$ and $b$.