How do you find the coordinates of the vertices, foci, and the equation of the asymptotes for the hyperbola $y^2/36-x^2/4=1$ ?

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Answer 1 · 2018-04-01T05:38:26

$"Vertices : "(0,+-6)," Asymptotes : "x+-3y=0, and,$

$Focii : (0,+-2sqrt10)$ .

We know that, for the $"Hyperbola S : "y^2/b^2-x^2/a^2=1$ ,

$(1) :" Vertices are "(0,+-b), (2)" Focii are "(0,+-be)$

$(3) :" the Asymptotes are "y=+-b/a*x$ .

The Eccentricity $e$ is given by, $a^2=b^2(e^2-1)$ .

We have, $b=6, a=2$ .

$:.$ Vertices are $(0,+-6)$ , and the eqns. of the Asymptotes are

$x+-3y=0$ .

For $e$ , we have, $4=36(e^2-1), or, e=sqrt10/3$ .

$:.$ Focii are $(0,+-2sqrt10)$ .

How do you find the coordinates of the vertices, foci, and the equation of the asymptotes for the hyperbola y^2/36-x^2/4=1y^2/36-x^2/4=1?