Problem 57 - Precalculus: Functions and Graphs

Question

Find an equation for the indicated part of the hyperbola. Lower branch of \(\frac{y^{2}}{25}-\frac{x^{2}}{36}=1\)

Step-by-Step Solution

Answer

The equation for the lower branch is \( y = - \sqrt{25 + \frac{25x^2}{36}} \).

Step 1: Identify the structure of the hyperbola

The given equation \( \frac{y^2}{25} - \frac{x^2}{36} = 1 \) represents a vertical hyperbola. This identifies the hyperbola as having a vertical transverse axis since \( y^2 \) is in the leading term.

Step 2: Determine center and orientation

For the standard form \( \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \), the center is at \((h, k)\). In this equation, the center is at \((0,0)\) because there are no shifted terms \((h,k)\). It's vertically oriented due to the \(y^2\) term leading.

Step 3: Solve for \(y\) to find the branches

From \( \frac{y^2}{25} - \frac{x^2}{36} = 1 \), solve for \( y^2 \) by isolating it: \( y^2 = 25 + \frac{25x^2}{36} \). Then, take the square root: \( y = \pm \sqrt{25 + \frac{25x^2}{36}} \).

Step 4: Extract the lower branch

The lower branch of the hyperbola is where \( y \) is negative. Thus, the equation for the lower branch is \( y = - \sqrt{25 + \frac{25x^2}{36}} \).

Key Concepts

Vertical Hyperbola

A hyperbola is a type of conic section that appears as two separate curves, called branches. A vertical hyperbola has a distinct characteristic: its transverse axis, or major axis, is aligned with the vertical y-axis. This means the y-term appears first in its standard equation.
For the equation \( \frac{y^2}{25} - \frac{x^2}{36} = 1 \), the position of the y variable in the leading term indicates that we're dealing with a vertical hyperbola. The transverse axis here is the vertical line where the hyperbola appears to stretch across vertically. This distinguishes it from a horizontal hyperbola, where the x-term would be in the leading position.

Hyperbola Branches

In hyperbolas, branches refer to the two distinct and separate curves that appear on each side of the center. These branches are symmetrical with respect to the center of the hyperbola.
In our example, the equation \( \frac{y^2}{25} - \frac{x^2}{36} = 1 \) creates two branches: an upper and a lower one. Each branch corresponds to where the hyperbola crosses vertically through the center point. To solve for exact branch equations, isolate the y variable and apply the square root, yielding two solutions for y: one positive for the upper branch and one negative for the lower.

• The positive square root represents the upper branch.
• The negative square root represents the lower branch.

The interest in this exercise is particularly the lower branch, given by \( y = - \sqrt{25 + \frac{25x^2}{36}} \).

Transverse Axis

The transverse axis of a hyperbola is its main axis, where the two branches appear to extend. For a vertical hyperbola, this axis is aligned with the y-axis.
This axis is significant because it provides information about the orientation and scale of the hyperbola. In our equation \( \frac{y^2}{25} - \frac{x^2}{36} = 1 \), the transverse axis implies that the hyperbola stretches vertically through the distance of \( \pm a \) from the center, where \( a^2 = 25 \). Thus, the transverse axis length is \( 2a = 10 \), spanning from \( y = -5 \) to \( y = 5 \).

Center of Hyperbola

The center of a hyperbola is the point equidistant from its vertices and from which the hyperbola's branches extend. In the standard form \( \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \), the center occurs at \((h, k)\).
Our equation \( \frac{y^2}{25} - \frac{x^2}{36} = 1 \) shows no horizontal or vertical shifts; thus, the center is at \((0, 0)\). This symmetry center is crucial for graphing and understanding the hyperbola's geometric properties. The center aligns with the origin, making it straightforward to see the symmetry across both x and y axes.