Ch. 7 - Conic Sections

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Blitzer 8th Edition

Ch. 7 - Conic Sections

Problema 90

Capitolo 8, Problema 90

Find the standard form of the equation of the hyperbola with vertices (5, −6) and (5, 6), passing through (0, 9).

Guida verificata passo dopo passo

Identify the orientation of the hyperbola by examining the vertices. Since the vertices are at (0, −6) and (0, 6), the hyperbola opens vertically along the y-axis.

Determine the center of the hyperbola, which is the midpoint of the vertices. Calculate the midpoint using the formula:

= \(\left\)( \(\frac{x_1 + x_2}{2}\), \(\frac{y_1 + y_2}{2}\) \(\right\))

. Here, the center is at (0, 0).

Find the distance between the center and each vertex, which gives the value of

a

. Since the vertices are at (0, ±6),

a = 6

Write the standard form of the equation of a vertical hyperbola centered at (0, 0):

\(\frac{y^2}{a^2}\) - \(\frac{x^2}{b^2}\) = 1

. Substitute

a = 6

to get

\(\frac{y^2}{36}\) - \(\frac{x^2}{b^2}\) = 1

Use the point (0, 9) that lies on the hyperbola to find

b^2

. Substitute

x = 0

and

y = 9

into the equation and solve for

b^2

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Standard Form of a Hyperbola

The standard form of a hyperbola depends on its orientation. For a vertical transverse axis centered at the origin, the equation is \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \). This form helps identify the vertices, foci, and asymptotes of the hyperbola.

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5:50

Asymptotes of Hyperbolas

Vertices and the Parameter \(a\)

Vertices are points where the hyperbola intersects its transverse axis. The distance from the center to each vertex is \(a\), so if vertices are at (0, −6) and (0, 6), then \(a = 6\). This value is crucial for writing the hyperbola's equation.

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5:30

Foci and Vertices of an Ellipse

Using a Point to Find \(b^2\)

Substituting a known point on the hyperbola into the standard form allows solving for \(b^2\). Given the point (0, 9), plugging it into the equation helps determine \(b^2\), completing the equation of the hyperbola.

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