Problem 40 - Algebra 2

Question

CHALLENGE A hyperbola with a horizontal transverse axis contains the point at \((4,3) .\) The equations of the asymptotes are \(y-x=1\) and \(y+x=5\) Write the equation for the hyperbola.

Step-by-Step Solution

Answer

The equation of the hyperbola is \( \frac{{(x-2)^2}}{4} - \frac{{(y-3)^2}}{4} = 1 \).

Step 1: Understand the standard form of a hyperbola equation

For the hyperbola with a horizontal transverse axis, the standard form is \( \frac{{(x-h)^2}}{a^2} - \frac{{(y-k)^2}}{b^2} = 1 \), where \((h, k)\) is the center of the hyperbola.

Step 2: Determine the slopes of the asymptotes

The equations for the asymptotes given are \( y-x=1 \) and \( y+x=5 \), which can be rewritten as \( y = x + 1 \) and \( y = -x + 5 \). Thus, the slopes of the asymptotes are \( 1 \) and \( -1 \). This indicates the slopes \( \pm \frac{b}{a} \) are equal to \( \pm 1 \), implying \( a = b \).

Step 3: Find the center of the hyperbola

The center \((h, k)\) of the hyperbola is the intersection point of the asymptotes. Solving the system of equations \( y = x + 1 \) and \( y = -x + 5 \), we get \( x = 2 \) and substituting \( x = 2 \) into \( y = x + 1 \) gives \( y = 3 \). Thus, the center is \((2, 3)\).

Step 4: Use the point on the hyperbola

The point \((4, 3)\) lies on the hyperbola. Substitute \( x = 4 \), \( y = 3 \), \( h = 2 \), and \( k = 3 \) into the hyperbola equation to find \( a \):\[ \frac{{(4-2)^2}}{a^2} - \frac{{(3-3)^2}}{a^2} = 1 \].This simplifies to \( \frac{4}{a^2} = 1 \), giving \( a^2 = 4 \) and \( a = 2 \). Since \( a = b \), \( b = 2 \) as well.

Step 5: Write the equation of the hyperbola

With \( h = 2 \), \( k = 3 \), and \( a = b = 2 \), substitute these into the standard hyperbola form:\[ \frac{{(x-2)^2}}{4} - \frac{{(y-3)^2}}{4} = 1 \].This is the equation of the hyperbola.

Key Concepts

Asymptotes

In the context of a hyperbola, asymptotes are lines that the hyperbola approaches but never actually meets. These lines give the hyperbola its distinctive shape and guide its curvature. For any hyperbola centered at \((h, k)\), the equations of the asymptotes are crucial in defining the geometry of the curve.

• Asymptotes cross at the center of the hyperbola.
• They represent the direction in which the hyperbola opens out.

In the provided exercise, the equations of the asymptotes are \(y - x = 1\) and \(y + x = 5\). Rewriting these equations as \(y = x + 1\) and \(y = -x + 5\) shows us the slopes of the asymptotes, which are 1 and -1. Because the absolute values of the slopes are equal, the transverse and conjugate axes have the same combined influence on the shape of the hyperbola, giving it a symmetrical appearance.

Transverse Axis

The transverse axis of a hyperbola is a central concept that defines the primary direction of the curve. Essentially, it acts as the backbone along which the two branches of the hyperbola open and extend.

• The transverse axis contains the vertices of the hyperbola.
• It is parallel to the line segment that joins the vertices.

In a hyperbola with a horizontal transverse axis, as seen in this exercise, the standard form is\[\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\]Here, the transverse axis runs along the x-direction. The point \((4, 3)\) lies on this axis when the hyperbola is centered at \((h, k) = (2, 3)\). This characteristic ensures that as \(x\) increases or decreases, \(y\) approaches the values dictated by the asymptotic lines. The length of the transverse axis is directly related to the value of \(a\), where \(2a\) is the total length between the vertices.

Center of the Hyperbola

The center of a hyperbola is a critical point that serves as the intersection of its asymptotes. It's essentially the balancing point or the origin around which the hyperbola is symmetrically laid out.

• The center is given by the point \((h, k)\).
• It's the point from which the distance to vertices and co-vertices is measured.

In the problem you're working on, by determining the intersection of the asymptotes \(y = x + 1\) and \(y = -x + 5\), you find the center to be \((2, 3)\). Solving these equations simultaneously, you calculate the coordinates of the center. This position helps us locate the hyperbola's symmetry and guides the setup of the entire equation, serving as the reference point for transformations and distances.