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$\\frac{x^2}{9}-\\frac{y^2}{4}=1$ the vertices of the hyperbola are $\\…

Question

$\frac{x^2}{9}-\frac{y^2}{4}=1$
the vertices of the hyperbola are
$\pm3, 0$
$\pm2, 0$
$(0, \pm3)$

Explanation:

Step1: Recall hyperbola vertex formula

For hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, vertices are $(\pm a, 0)$.

Step2: Identify $a$ from given equation

Given $\frac{x^2}{9}-\frac{y^2}{4}=1$, so $a^2=9$, thus $a=\sqrt{9}=3$.

Step3: Write vertex coordinates

Substitute $a=3$ into vertex form: $(\pm 3, 0)$.

Answer:

$(\pm 3, 0)$