Question

determine the center, vertices, and foci for the hyperbola.

$\frac{x^{2}}{16}-\frac{y^{2}}{36}=1$

enter your answers as points $(a,b)$. enter the vertices and foci in increasing order of the $x$-coordinate.

the center is

the vertices are and

the foci are and

Explanation:

Step1: Recall hyperbola standard - form

The standard form of a hyperbola centered at the origin is $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}} = 1$. For the given hyperbola $\frac{x^{2}}{16}-\frac{y^{2}}{36}=1$, we have $a^{2}=16$ and $b^{2}=36$.

Step2: Find the center

For a hyperbola of the form $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}} = 1$, the center is at the origin $(0,0)$.

Step3: Find the vertices

Since $a^{2}=16$, then $a = 4$. The vertices of the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}} = 1$ are $(\pm a,0)$. So the vertices are $(-4,0)$ and $(4,0)$.

Step4: Find the value of $c$

We know the relationship $c^{2}=a^{2}+b^{2}$. Substituting $a^{2}=16$ and $b^{2}=36$, we get $c^{2}=16 + 36=52$, so $c = 2\sqrt{13}$.

Step5: Find the foci

The foci of the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}} = 1$ are $(\pm c,0)$. So the foci are $(-2\sqrt{13},0)$ and $(2\sqrt{13},0)$.

Answer:

The center is $(0,0)$.
The vertices are $(-4,0)$ and $(4,0)$.
The foci are $(-2\sqrt{13},0)$ and $(2\sqrt{13},0)$.