hyperbolas

[lillie5455]

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
(x-h)^2/a^2 - (y-k)^2/b^2 or
(y-k)^2/a^2 - (x-h)^2/b^2

 

[soroban]

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: $\displaystyle \:\frac{x^2}{49}\,-\,\frac{y^2}{32}\;=\;1$