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$Find the vertices, the foci, and the equations of the asymptotes of the hyperbola. {:[(x^(2))/(36)-(y^(2))/(4)=1],[" vertex "(x","y)=(◻","0",")" (smaller "x"-value) "],[" vertex "(x","y)=(◻","0)" (larger "x"-value) "],[" focus "(x","y)=(◻","0)" (smaller "x"-value) "],[" focus "(x","y)=(◻","0)" (larger "x"-value) "],[" asymptote "y=◻" (smaller slope) "],[" asymptote "quad y=◻" (larger slope) "]:}$

Find the vertices, the foci, and the equations of the asymptotes of the hyperbola. $\left[\frac{x^{2}}{36}-\frac{y^{2}}{4}=1\right]$ , $\text{vertex }(x,y)=(\Box,0)$ (smaller $x$ -value), $\text{vertex }(x,y)=(\Box,0)$ (larger $x$ -value), $\text{focus }(x,y)=(\Box,0)$ (smaller $x$ -value), $\text{focus }(x,y)=(\Box,0)$ (larger $x$ -value), $\text{asymptote }y=\Box$ (smaller slope), $\text{vertex }(x,y)=(\Box,0)$ $0$ (larger slope).

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Math Problems

Algebra 2

Find the focus or directrix of a parabola

Full solution

Q. Find the vertices, the foci, and the equations of the asymptotes of the hyperbola.

\left[\frac{x^{2}}{36}-\frac{y^{2}}{4}=1\right]

\text{vertex }(x,y)=(\Box,0)

(smaller

x

-value),

\text{vertex }(x,y)=(\Box,0)

(larger

x

-value),

\text{focus }(x,y)=(\Box,0)

(smaller

x

-value),

\text{focus }(x,y)=(\Box,0)

(larger

x

-value),

\text{asymptote }y=\Box

(smaller slope),

\text{vertex }(x,y)=(\Box,0)

0

(larger slope).

Identify Hyperbola Parameters: Identify the standard form of the hyperbola and its parameters. $\newline$ Given equation: $\frac{x^2}{36} - \frac{y^2}{4} = 1$ . $\newline$ This is a horizontal hyperbola because the $x^2$ term is positive. $\newline$ Here, $a^2 = 36$ and $b^2 = 4$ , so $a = 6$ and $b = 2$ .
Calculate Vertices: Calculate the vertices of the hyperbola. $\newline$ Vertices are at $(\pm a, 0)$ for a horizontal hyperbola. $\newline$ Substitute $a = 6$ : vertices are at $(\pm 6, 0)$ , which are $(6, 0)$ and $(-6, 0)$ .
Calculate Foci: Calculate the foci of the hyperbola. $\newline$ Foci are at $(\pm c, 0)$ , where $c = \sqrt{a^2 + b^2}$ . $\newline$ Calculate $c = \sqrt{36 + 4} = \sqrt{40} = 2\sqrt{10}$ . $\newline$ Foci are at $(\pm 2\sqrt{10}, 0)$ , which are $(2\sqrt{10}, 0)$ and $(-2\sqrt{10}, 0)$ .
Derive Asymptote Equations: Derive the equations of the asymptotes for the hyperbola. $\newline$ Asymptotes for a horizontal hyperbola are given by $y = \pm \frac{b}{a}x$ . $\newline$ Substitute $b = 2$ and $a = 6$ : $y = \pm \frac{1}{3}x$ . $\newline$ Equations of the asymptotes are $y = \frac{1}{3}x$ and $y = -\frac{1}{3}x$ .