How do you find the center, vertices, foci and asymptotes of x^2/7 - y^2/9=1?

1 Answer
Parabola
Dec 21, 2017

The graph should look like this:
graph{x^2/7-y^2/9=1 [-10, 10, -5, 5]}

Explanation:

x^2/7-y^2/9=1

Since we are subtracting, we know that this is a hyperbola. Also, since the fraction with x^2 is positive, this hyperbola opens to the right and the left.
Therefore, equation is in the form x^2/a^2-y^2/b^2
We know that:
h=0
k=0
a=sqrt7
b=sqrt9 or 3.
c=sqrt (a^2+b^2) or 4.
Knowing this, we already know which equations to use:
The center is always (h,k)
The vertices are (h+a,k) and (h-a,k)
The foci are (h+c,k) and (h-c,k)
Since this hyperbola opens sideways, the asymptotes are found by this formula:
y=k+-b/a(x-h)

Therefore, we know that the center is (0,0)
The vertices are approximately (2.645,0) and (-2.645,0)
The foci are (4,0) and (-4,0)
The lines of asymptotes are y=(3sqrt7)/7x and y=-(3sqrt7)/7x

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