Question

find the vertices of the hyperbola. enter the smallest x coordinate first. $\frac{x^{2}}{9}-\frac{y^{2}}{49}=1$

Explanation:

Step1: Identify the standard - form of hyperbola

The standard form of a horizontal hyperbola is $\frac{(x - h)^2}{a^2}-\frac{(y - k)^2}{b^2}=1$, and its vertices are $(h\pm a,k)$. For the given hyperbola $\frac{x^{2}}{9}-\frac{y^{2}}{49}=1$, we have $h = 0,k = 0,a^{2}=9,b^{2}=49$.

Step2: Calculate the value of a

Since $a^{2}=9$, then $a = 3$ (we take the positive square - root as $a$ represents a distance).

Step3: Find the vertices

The vertices of the hyperbola $\frac{x^{2}}{9}-\frac{y^{2}}{49}=1$ are $(h - a,k)$ and $(h + a,k)$. Substituting $h = 0,k = 0,a = 3$, we get the vertices $(-3,0)$ and $(3,0)$.

Answer:

$(-3,0)$ and $(3,0)$