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Chapter 6: Problem 63

Consider a hyperbola centered at the origin with a horizontal transverse axis. Use the definition of a hyperbola to derive its standard form.

Short Answer

Expert verified
The equation of the hyperbola centered at the origin with a horizontal transverse axis is \(x^2 / a^2 - y^2 / b^2 = 1\).

Step by step solution

01

Deriving Distance Formula of a Hyperbola

The definition of a hyperbola is that for any point (x, y) on the hyperbola, the difference between the distances to the two foci is constant and equals 2a. If we consider the foci to be at positions (c, 0) and (-c, 0), then the distance formula to each focus from a point (x, y) on the hyperbola is given by \(d_1 = \sqrt{(x - c)^2 + y^2}\) and \(d_2 = \sqrt{(x + c)^2 + y^2}\). So, we have \( |d_1 - d_2| = 2a \).
02

Substituting the Distances into the Definition

Substituting the distances into the definition of a hyperbola gives \(\left|\sqrt{(x - c)^2 + y^2} - \sqrt{(x + c)^2 + y^2}\right| = 2a \). Since the difference in distance can be negative, there are two cases to consider.
03

Simplify the equation

For the first case \(\sqrt{(x + c)^2 + y^2} - \sqrt{(x - c)^2 + y^2} = 2a \). Square both sides to eliminate the square roots will result in \((x + c)^2 + y^2 = 4a^2 + 4acx + (x - c)^2 + y^2 \). The y^2 terms cancel out, and simplifying give us \(x^2 - a^2 = c^2 \). Using the Pythagorean relationship, we can substitute for c^2 to give \(x^2 - a^2 = a^2 + b^2 \), which simplifies to \(x^2 / a^2 - y^2 / b^2 = 1\). Similarly, the second case yields \(y^2 / b^2 - x^2 / a^2 = 1\).
04

Final Equation

The two cases y^2 / b^2 - x^2 / a^2 = 1 and x^2 / a^2 - y^2 / b^2 = 1 represent the same hyperbola. The first case corresponds to a vertical transverse axis and the second to a horizontal transverse axis. Because it was stated that the hyperbola has a horizontal transverse axis, the final equation for this hyperbola is \(x^2 / a^2 - y^2 / b^2 = 1\)

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