![\bf \textit{hyperbolas, vertical traverse axis } \\\\ \cfrac{(y- k)^2}{ a^2}-\cfrac{(x- h)^2}{ b^2}=1 \qquad \begin{cases} center\ ( h, k)\\ vertices\ ( h, k\pm a)\\ c=\textit{distance from}\\ \qquad \textit{center to foci}\\ \qquad \sqrt{ a ^2 + b ^2}\\ asymptotes\quad y= k\pm \cfrac{a}{b}(x- h) \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}](/tpl/images/0337/2587/fb032.png)
![\cfrac{(y-3)^2}{1}-\cfrac{(x+2)^2}{4}=1\implies \cfrac{[y-3]^2}{1^2}-\cfrac{[x-(-2)]^2}{2^2}=1~~ \begin{cases} h=-2\\ k=3\\ a=1\\ b=2 \end{cases} \\\\[-0.35em] ~\dotfill\\\\ c=\sqrt{a^2+b^2}\implies c=\sqrt{1+4}\implies c=\sqrt{5} \\\\\\ \stackrel{\textit{so then the foci are at}}{(-2~~,~~3\pm \sqrt{5})}\qquad \qquad \qquad \stackrel{\textit{and its vertices are at }}{(-2~~,~~3\pm 1)}\implies \begin{cases} (-2,4)\\ (-2,2) \end{cases}](/tpl/images/0337/2587/72832.png)
![\bf y=3\pm \cfrac{1}{2}[x-(-2)]\implies y=3\pm \cfrac{1}{2}(x+2) \\\\[-0.35em] ~\dotfill\\\\ y=3+ \cfrac{1}{2}(x+2)\implies y=3+\cfrac{x+2}{2}\implies y=\cfrac{6+x+2}{2} \\\\\\ y=\cfrac{x+8}{2}\implies y=\cfrac{1}{2}x+4 \\\\[-0.35em] ~\dotfill\\\\ y=3- \cfrac{1}{2}(x+2)\implies y=3-\cfrac{(x+2)}{2}\implies y=\cfrac{6-(x+2)}{2} \\\\\\ y=\cfrac{6-x-2}{2}\implies y=\cfrac{-x+4}{2}\implies y=-\cfrac{1}{2}x+2](/tpl/images/0337/2587/21f47.png)
Hyperbolas
label the foci, the vertices, and the asymptotes.
(y−3)^2/1 - (x+...
Mathematics, 20.10.2019 14:30 ronaldotheexplorer12
Hyperbolas
label the foci, the vertices, and the asymptotes.
(y−3)^2/1 - (x+2)^2/4 = 1
Answers: 1
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