Рівняння гіперболи та асимптоти

Problem

Find the standard form equation, the asymptotes, and the eccentricity of the hyperbola with vertices at $(-3,0)$ and $(3,0)$ and foci at $(-5,0)$ and $(5,0)$.

Step 1: Use the vertices and foci to get $a$ and $c$

Since the vertices and foci are both on the $x$-axis, the hyperbola has center at the origin and opens left-right, so its standard form is

$$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1.$$

From the vertices, $a = 3$. From the foci, $c = 5$.

Using the hyperbola relation

$$c^2 = a^2 + b^2,$$

we get

$$25 = 9 + b^2,$$

so

$$b^2 = 16.$$

Step 2: Write the hyperbola equation

Substituting $a^2 = 9$ and $b^2 = 16$ into the standard form gives

$$\frac{x^2}{9} - \frac{y^2}{16} = 1.$$

Step 3: Find the asymptotes

For a hyperbola of this form, the asymptotes are

$$y = \pm \frac{b}{a}x.$$

With $a = 3$ and $b = 4$,

$$y = \pm \frac{4}{3}x.$$

Step 4: Compute the eccentricity

The eccentricity is

$$e = \frac{c}{a} = \frac{5}{3}.$$

Answer

The hyperbola is $\dfrac{x^2}{9} - \dfrac{y^2}{16} = 1$, its asymptotes are $y = \pm \dfrac{4}{3}x$, and its eccentricity is $\dfrac{5}{3}$.