Mathematics, 04.07.2019 06:00 lhmsokol56
Hello! could anyone me with this question? i don't understand it. question: the equation x^2/24^2-y^/--^2 represents a hyperbola centered at the origin with a directix of x=576/26. the positive value correctly fills in the blank in the equation. i'll give 50 points for this question ^^
Answers: 1
![\bf \textit{hyperbolas, horizontal traverse axis } \\\\ \cfrac{(x- h)^2}{ a^2}-\cfrac{(y- k)^2}{ b^2}=1 \qquad \begin{cases} center\ ( h, k)\\ vertices\ ( h\pm a, k)\\ c=\textit{distance from}\\ \qquad \textit{center to foci}\\ \qquad \sqrt{ a ^2 + b ^2}\\ \textit{asymptotes}\quad y= k\pm \cfrac{b}{a}(x- h)\\[0.8em] directrix\qquad x=\cfrac{a^2}{c} \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}](/tpl/images/0049/1937/42d87.png)
![\bf \cfrac{x^2}{24^2}-\cfrac{y^2}{b^2}=1\implies \cfrac{(x-0)^2}{24^2}-\cfrac{(y-0)^2}{b^2}=1 \\\\\\ \stackrel{\textit{we know that}}{x=\cfrac{a^2}{c}}\implies \cfrac{576}{26}=\cfrac{24^2}{c}\implies \cfrac{576}{26}=\cfrac{576}{c}\implies c=\cfrac{576\cdot 26}{576}\implies \boxed{c=26} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \stackrel{\textit{we also know that}}{c=\sqrt{a^2+b^2}}\implies c^2-a^2=b^2\implies \sqrt{c^2-a^2}=b~~ \begin{cases} c=26\\ a=24 \end{cases}](/tpl/images/0049/1937/4831d.png)
![\bf \sqrt{26^2-24^2}=b\implies \sqrt{100}=b\implies \boxed{10=b} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill \cfrac{x^2}{24^2}-\cfrac{y^2}{10^2}=1~\hfill](/tpl/images/0049/1937/6fb1c.png)
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Hello! could anyone me with this question? i don't understand it. question: the equation x^2/24^...
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