How do you find the coordinates of the vertices, foci, and the equation of the asymptotes for the hyperbola $x^2/81-y^2/49=1$ ?

Question

3683 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-20T14:41:29

The vertices are $(9,0)$ and $(-9,0)$
The foci are F $(sqrt130,0)$ and F' $(-sqrt130,0)$
The asymptotes are $y=7/9x$ and $y=-7/9x$

Let's compare this equation to the gerenal equation of a left-right hyperbola

$(x-h)^2/a^2-(y-k)^2/b^2=1$

$x^2/81-y^2/49=1$

The center is $(h,k)=(0,0)$

The vertices are $(h+-a,k)=(+-9,0)$

The slope of the asymptotes are $+-b/a=+-7/9$

The equations of the asymptotes are $y=b/a(x-h)$

The asymptotes are $y=7/9x$ and $y=-7/9x$

To determine the foci, we need $c=+-sqrt(a^2+b^2)$

$c=+-sqrt(81+49)=sqrt130$

The foci are F $=(h+c,k)=(sqrt130,0)$

and F' $=(h-c,k)=(-sqrt130,0)$

graph{(x^2/81-y^2/49-1)(y-7/9x)(y+7/9x)=0 [-20.27, 20.29, -10.14, 10.13]}

How do you find the coordinates of the vertices, foci, and the equation of the asymptotes for the hyperbola x^2/81-y^2/49=1x^2/81-y^2/49=1?