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Find the vertices of the hyperbola. Enter the smallest coordinate first.

[tex]\[
\frac{x^2}{9} - \frac{y^2}{49} = 1
\][/tex]

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Sagot :

To find the vertices of the hyperbola given by the equation
[tex]\[ \frac{x^2}{9} - \frac{y^2}{49} = 1, \][/tex]
we start by recognizing the standard form of a hyperbola with a horizontal transverse axis. The standard form is:
[tex]\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1. \][/tex]

In our equation, we can identify [tex]\( a^2 \)[/tex] and [tex]\( b^2 \)[/tex]:
[tex]\[ a^2 = 9 \quad \text{and} \quad b^2 = 49. \][/tex]

To find [tex]\( a \)[/tex] and [tex]\( b \)[/tex], we take the square roots of [tex]\( a^2 \)[/tex] and [tex]\( b^2 \)[/tex]:
[tex]\[ a = \sqrt{9} = 3 \quad \text{and} \quad b = \sqrt{49} = 7. \][/tex]

The vertices of a hyperbola with a horizontal transverse axis are located at [tex]\( (\pm a, 0) \)[/tex]. Thus, for our hyperbola:
[tex]\[ \text{Vertices are at } (-a, 0) \text{ and } (a, 0). \][/tex]

Substituting [tex]\( a = 3 \)[/tex]:
[tex]\[ \text{Vertices are at } (-3, 0) \text{ and } (3, 0). \][/tex]

To follow the instruction to enter the smallest coordinate first, we order these coordinates correctly:
[tex]\[ ((-3, 0), (3, 0)). \][/tex]

Therefore, the vertices of the hyperbola are:
[tex]\[ ((-3, 0), (3, 0)). \][/tex]