Question

Find the equation of the given central conic. Hyperbola whose asymptotes are \(x \pm 2 y=0\) and that goes through the point \((4,3)\)

Short answer

The hyperbola equation is \( \frac{4x^2}{55} - \frac{y^2}{55} = 1 \).

Step-by-step solution

Identify the form of the hyperbola equation

Given the equations of the asymptotes, which are \( x \pm 2y = 0 \), we know these have slopes of \( m_1 = 2 \) and \( m_2 = -2 \). We can deduce that this hyperbola is centered at the origin and has the equation of the form \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) or \( \frac{y^2}{b^2} - \frac{x^2}{a^2} = 1 \).

Relate slopes to the hyperbola equation

The slopes of the asymptotes for a hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) are given by \( \pm \frac{b}{a} \). Since the slopes are \( \pm 2 \), we equate \( \frac{b}{a} = 2 \).

Solve for variable relationship

From the relationship \( \frac{b}{a} = 2 \), cross-multiply to find \( b = 2a \).

Substitute known point into the equation

Using the equation form \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), we substitute the point (4, 3) into the equation: \( \frac{4^2}{a^2} - \frac{3^2}{(2a)^2} = 1 \).

Simplify the equation

Simplifying gives: \( \frac{16}{a^2} - \frac{9}{4a^2} = 1 \). Make the denominators equal: \( \frac{64}{4a^2} - \frac{9}{4a^2} = 1 \), resulting in \( \frac{55}{4a^2} = 1 \).

Solve for \(a^2\)

Rearranging gives \( 4a^2 = 55 \), so \( a^2 = \frac{55}{4} \).

Determine \(b^2\) using \(b = 2a\)

From \( b = 2a \), \( b^2 = 4a^2 = 55 \).

Write the final hyperbola equation

Substitute values back into the equation \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) to get \( \frac{x^2}{\frac{55}{4}} - \frac{y^2}{55} = 1 \). It simplifies to the equation \( \frac{4x^2}{55} - \frac{y^2}{55} = 1 \).

Key concepts

Asymptotes

Asymptotes are lines that a curve approaches as it heads towards infinity. For hyperbolas, asymptotes offer crucial insight into their structure and orientation. The hyperbola's arms grow closer and closer to these lines but never actually intersect them.
In the original exercise, the given asymptotes are described by the equations \( x + 2y = 0 \) and \( x - 2y = 0 \). These lines have slopes of \( m_1 = 2 \) and \( m_2 = -2 \) and intersect at the origin.

• For hyperbolas of the form \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), the slopes of asymptotes are given by \( \pm \frac{b}{a} \).
• From the problem, since \( \pm \frac{b}{a} = \pm 2 \), we can identify \( b = 2a \).

Understanding these asymptotes helps in shaping the hyperbola properly. They guide us to conclude that the central hyperbola is centered at the origin.

Central Conic

A central conic is a conic section, such as a circle, ellipse, or hyperbola, that is symmetrically centered around a point, typically the origin.
This symmetry allows us to derive significant geometric properties based on the central position.

• In the case of hyperbolas, an equation like \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) indicates a central conic centered at the origin (0,0).
• The given task shows a hyperbola centered at the origin with given asymptotes.

Recognizing hyperbolas as central conics helps us define equations better and relate the coefficients to spatial characteristics.

Conic Sections

Conic sections are produced by intersecting a plane with a double-napped cone. Depending on the angle of the plane's intersection, different conic sections emerge. These include circles, ellipses, parabolas, and hyperbolas.
Hyperbolas arise when the intersecting plane cuts both naps of the cone. This results in two separate curves that face away from the central point.

• Hyperbolas can be represented in standard form equations such as \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \).
• Central conics are a fundamental subset of conic sections, providing symmetry and easy calculation for points, vertices, and asymptotes.

In the task above, the identification of asymptotes sets a clear understanding of the hyperbola's orientation and lets us determine the central conic equation that represents the curve accurately.