Hello, lillie5455!
I must assume you know
something about hyperbolas . . .
i don't understand how to write the equation of a parabola ?
with vertices at (-7,0) and (7,0) and foci at (-9,0) and (9,0)
I have to use the equation: \(\displaystyle \,\frac{(x-h)^2}{a^2}\,-\,\frac{(y-k)^2}{b^2}\:=\:1\,\) or \(\displaystyle \,\frac{(y-k)^2}{a^2}\,-\,\frac{(x-h)^2}{b^2}\)
To use those equations, we need to know the center \(\displaystyle (h,k)\) and the values of \(\displaystyle a\) and \(\displaystyle b.\)
Since the vertices are symmetric about the origin, the center is \(\displaystyle (h,k)\,-\,(0,0).\)
Since the vertices are oriented "horizontally", we use the first form.
Since the vertices are \(\displaystyle (\pm7,0)\), we have: \(\displaystyle \,a\,=\,7\)
The foci are \(\displaystyle (\pm9,0)\), so we have: \(\displaystyle \,c\,=\,9\)
For hyperbolas, the focal equation is: \(\displaystyle \.c^2\:=\:a^2\,-\,b^2\)
\(\displaystyle \;\;\)So we have: \(\displaystyle \,9^2\:=\:7^2\,+\,b^2\;\;\Rightarrow\;\;b^2\,=\,32\)
And we have the equation: \(\displaystyle \:\frac{x^2}{49}\,-\,\frac{y^2}{32}\;=\;1\)
