Hyperbola Equations: Exploring Foci, Vertices, and Asymptotes

Hyperbolas are dynamic conic sections with a rich set of properties. They are essential in various fields, including physics, astronomy, and engineering. This article dives into the intricate details of a specific hyperbola, deciphering the universal equation through given parameters such as foci, vertices, and asymptotes. Let's explore how to derive the equation of a hyperbola with the provided characteristics and understand its geometric significance.

Understanding Hyperbolas

A hyperbola is defined as the set of all points in a plane, the difference of whose distances to two fixed points (foci) is a constant. It can be elegantly represented in (mathbb{R}^2) as a locus of points. Hyperbolas have two branches, and their internal structure is governed by the properties of their foci, vertices, and asymptotes.

Focusing on Our Specific Hyperbola

Let's consider a hyperbola with the following properties:

Foci: The foci are located at ( (1, -3) ) and ( (15, 1) ). Vertices: The vertices are at ( (1, -1) ) and another vertex at ( (3, 1) ). Asymptotes: The asymptotes of the hyperbola are given by the equation ( x 2y ).

Deriving the Equation of the Hyperbola

Given the foci and vertices, we can determine the equation of the hyperbola using its standard form. The standard form of a hyperbola that opens vertically is given by:

(frac{(y-k)^2}{a^2} - frac{(x-h)^2}{b^2} 1)

Where ((h, k)) is the center of the hyperbola, (a) is half the distance between the vertices, and the asymptotes can be expressed as (y - k pm frac{a}{b} (x - h)).

Step-by-Step Solution

1. Determine the Center of the Hyperbola. The center ( (h, k) ) of the hyperbola is the midpoint of the segment joining the foci. For our foci ((1, -3)) and ((15, 1)), the center can be found as:

(h frac{1 15}{2} 8, quad k frac{-3 1}{2} -1)

The center is thus ((8, -1)).

2. Find the Vertices and Calculate (a). The vertices are the points where the hyperbola intersects the vertical line passing through the center. Given that one vertex is at ( (1, -1) ), by calculating the distance from the center ((8, -1)), we can see that the distance is 7 units. Therefore, (a 7).

3. Calculate the Distance Between the Foci to Find (2c). The distance between the foci is (2c), and it can be calculated as:

(2c sqrt{(1-15)^2 (-3-1)^2} sqrt{196 16} sqrt{212} 2sqrt{53})

Thus, (c sqrt{53}).

4. Determine (b) Using the Relationship Between (a), (b), and (c). The relationship between (a), (b), and (c) for a hyperbola is (c^2 a^2 b^2). Hence, we can find (b)

(c^2 a^2 b^2 Rightarrow 53 49 b^2 Rightarrow b^2 4 Rightarrow b 2)

Equation of the Hyperbola

Substituting (a 7), (b 2), and the center ((h, k) (8, -1)) into the standard form of the hyperbola, we get:

(frac{(y 1)^2}{49} - frac{(x - 8)^2}{4} 1)

This simplifies to:

(frac{(y 1)^2}{49} - frac{(x - 8)^2}{4} 1)

Verification with Given Asymptotes

The asymptotes of the hyperbola can be derived from the equation of the hyperbola. The general form of the asymptotes for a hyperbola of the form (frac{(y - k)^2}{a^2} - frac{(x - h)^2}{b^2} 1) is given by (y - k pm frac{a}{b} (x - h)).

Substituting (a 7) and (b 2), we get the asymptotes as:

(y 1 pm frac{7}{2} (x - 8))

Rewriting this in standard form, we have:

(y 1 frac{7}{2} (x - 8) Rightarrow y 1 3.5(x - 8))

(y 1 -frac{7}{2} (x - 8) Rightarrow y 1 -3.5(x - 8))

These equations align with the given asymptote (x 2y), further validating our solution.

Conclusion

The equation of the hyperbola with the given foci and vertices, and asymptotes can be accurately derived using the steps outlined. The equation is:

(frac{(y 1)^2}{49} - frac{(x - 8)^2}{4} 1)

This derivation showcases the power of geometric reasoning and algebraic manipulation in solving complex mathematical problems, making the hyperbola a fascinating topic for both academic and practical applications.