How do you find the center, vertices, foci, and asymptotes of the hyperbola, and sketch its graph of $x^2/9 - y^2/16 = 1$ ?

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Answer 1 · 2017-03-04T15:31:40

The center is $(0,0)$
The vertices are (-3,0) $and$ (3,0)#

The foci are $F'=(-5,0)$ and $F=(5,0)$
The asymptotes are $y=4/3x$ and $y=-4/3x$

We compare this equation

$x^2/3^2-y^2/4^2=1$

to

$x^2/a^2-y^2/b^2=1$

The center is $C=(0,0)$

The vertices are $V'=(-a,0)=(-3,0)$ and $V=(a,0)=(3,0)$

To find the foci, we need the distance from the center to the foci

$c^2=a^2+b^2=9+16=25$

$c=+-5$

The foci are $F'=(-c,0)=(-5,0)$ and $F=(c,0)=(5,0)$

$x^2/3^2-y^2/4^2=0$

$y=+-4/3x$

graph{((x^2)/9-(y^2)/16-1)(y-4/3x)(y+4/3x)=0 [-16.02, 16.02, -8.01, 8.01]}

How do you find the center, vertices, foci, and asymptotes of the hyperbola, and sketch its graph of x^2/9 - y^2/16 = 1x^2/9 - y^2/16 = 1?