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Asymptotes $y = \pm \frac{5}{4}x$ one vertex is $(0,3)$ what is equation of hyperbola

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Find equations of tangent lines using limits

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Q. Asymptotes

y = \pm \frac{5}{4}x

one vertex is

(0,3)

what is equation of hyperbola

Identify Conic Section: Identify the type of conic section based on the given asymptotes and vertex. The asymptotes $y = \pm \frac{5}{4}x$ suggest a hyperbola centered at the origin, and the vertex at $(0,3)$ indicates a vertical hyperbola.
Standard Form: Use the standard form of a vertical hyperbola centered at the origin, which is $y-k)^2/a^2 - (x-h)^2/b^2 = 1\, where \$h,k$ is the center and ' $a$ ' is the distance from the center to each vertex along the y-axis.
Center and 'a': Since the vertex given is $(0,3)$ , and it's a vertical hyperbola, the center is at $(0,0)$ and ' $a$ ' is $3$ . This is because the vertex is $3$ units away from the center along the y-axis.
Solve for 'b': The slopes of the asymptotes for a vertical hyperbola are given by $\pm\frac{a}{b}$ . We know the slopes are $\pm\frac{5}{4}$ , so setting $\frac{a}{b} = \frac{5}{4}$ and knowing $a = 3$ , solve for 'b'. $\newline$ $\frac{a}{b} = \frac{5}{4}$ $\newline$ $\frac{3}{b} = \frac{5}{4}$ $\newline$ $b = 3 \times \frac{4}{5}$ $\newline$ $b = \frac{12}{5} \text{ or } 2.4$
Plug Values into Equation: Plug the values of $a$ and $b$ into the standard form equation of the hyperbola. $\newline$ $(y-0)^2/3^2 - (x-0)^2/(12/5)^2 = 1$ $\newline$ $y^2/9 - x^2/(2.88) = 1$

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