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Geometry

Hyperbola Equation and Asymptotes

A hyperbola has vertices at (±3, 0) and foci at (±5, 0). Find the standard form equation using c² = a² + b², determine the asymptote equations y = ±(b/a)x, and calculate the eccentricity.

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Problem

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

Step 1: Use the vertices and foci to get aa and cc

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

x2a2y2b2=1.\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1.

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

Using the hyperbola relation

c2=a2+b2,c^2 = a^2 + b^2,

we get

25=9+b2,25 = 9 + b^2,

so

b2=16.b^2 = 16.

Step 2: Write the hyperbola equation

Substituting a2=9a^2 = 9 and b2=16b^2 = 16 into the standard form gives

x29y216=1.\frac{x^2}{9} - \frac{y^2}{16} = 1.

Step 3: Find the asymptotes

For a hyperbola of this form, the asymptotes are

y=±bax.y = \pm \frac{b}{a}x.

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

y=±43x.y = \pm \frac{4}{3}x.

Step 4: Compute the eccentricity

The eccentricity is

e=ca=53.e = \frac{c}{a} = \frac{5}{3}.

Answer

The hyperbola is x29y216=1\dfrac{x^2}{9} - \dfrac{y^2}{16} = 1, its asymptotes are y=±43xy = \pm \dfrac{4}{3}x, and its eccentricity is 53\dfrac{5}{3}.

Concepts

Hyperbolas and Their Equations

A hyperbola is the set of all points whose distances to two fixed points (foci) differ by a constant. It has two branches, asymptotes, and can open horizontally or vertically.

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Hyperbola Equation and Asymptotes