Problem 32 - College Algebra with Corequisite Support

Question

For the following exercises, use any method to solve the nonlinear system. $$ \begin{aligned} 16 x^{2}-9 y^{2}+144 &=0 \\ y^{2}+x^{2} &=16 \end{aligned} $$

Step-by-Step Solution

Answer

The solutions are \( (0, 4) \) and \( (0, -4) \).

Step 1: Identify the Type of System

The given system is nonlinear and consists of a hyperbola given by \( 16x^2 - 9y^2 + 144 = 0 \) and a circle represented by \( y^2 + x^2 = 16 \). We need to find the points of intersection between these two figures.

Step 2: Simplify and Rearrange the Equations

First, rearrange the equation of the hyperbola to standard form. The equation \( 16x^2 - 9y^2 = -144 \) can be rewritten as \( \frac{x^2}{9} - \frac{y^2}{16} = -1 \). There's already the second equation \( x^2 + y^2 = 16 \) in its standard form.

Step 3: Substitute to Reduce the Equations

From the circle's equation \( y^2 = 16 - x^2 \). Substitute \( y^2 \) into the hyperbola equation. This substitution will eliminate \( y^2 \) in favor of \( x^2 \).

Step 4: Solve the Substituted Equation

Substitute \( y^2 = 16 - x^2 \) into the hyperbola equation: \( 16x^2 - 9(16 - x^2) + 144 = 0 \). Simplify it to get \( 25x^2 = 0 \).

Step 5: Solve for \( x \) and \( y \)

From \( 25x^2 = 0 \), we find \( x = 0 \). Substitute \( x = 0 \) back into the circle equation to find \( y \). So, substitute into \( x^2 + y^2 = 16 \) gives \( y^2 = 16 \), hence \( y = \pm 4 \).

Step 6: Verify Solutions

Check if the points \( (0, 4) \) and \( (0, -4) \) satisfy both original equations. Substitution into the hyperbola's equation: \( 16(0)^2 - 9(\pm 4)^2 + 144 = 0 \) and into the circle's equation: \( (0)^2 + (\pm 4)^2 = 16 \). Both equations are true.

Key Concepts

Hyperbola

A hyperbola is a type of conic section that consists of two separate curves, which are mirror images of each other. It can be visualized as the shape you get when you slice a cone with a plane. The general equation for a hyperbola is given by \[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \] where \(a\) and \(b\) are constants that determine the shape and size of the hyperbola.
In the context of the exercise, the hyperbola equation is \[ 16x^2 - 9y^2 + 144 = 0 \]. When you rearrange it into standard form as shown in the solution, you get \[ \frac{x^2}{9} - \frac{y^2}{16} = -1 \]. This particular form of hyperbola indicates that it opens horizontally along the x-axis.
Key characteristics of a hyperbola include:

• The center, which is the midpoint between its vertices.
• Asymptotes: Lines that the hyperbola approaches but never touches.
• Vertices: Points where the hyperbola intersects its principal axis.

Understanding these properties helps in sketching the curve and analyzing its intersections with other figures, such as the circle in our system.

Circle

A circle is another kind of conic section, characterized by all points being equidistant from a center point. The simplest form of a circle's equation is \[ (x - h)^2 + (y - k)^2 = r^2 \], where \( (h, k) \) is the center and \( r \) is the radius.
Here, the circle’s equation is \[ x^2 + y^2 = 16 \]. This indicates a circle centered at the origin \((0,0)\) with a radius of 4, since \( 16 \) equals \( 4^2 \). The circle's properties include:

• A central point, known as the center.
• A consistent radius from the center to the circumference.
• It is symmetrical in all directions from the center.

Circles are straightforward to handle because of their simplicity and symmetry, making it easier to combine their equations with others, like those of hyperbolas, to determine points of intersection.

System of Equations

A system of equations is a set of two or more equations with the same variables. In the case of nonlinear systems, these equations are not limited to straight lines but can involve curves such as circles or hyperbolas. Solving such a system means finding the point(s) where the equations intersect. It's like finding a common solution that satisfies all equations at once.
For the given problem, the nonlinear system comprised a hyperbola and a circle:

• \(16x^2 - 9y^2 + 144 = 0\) (hyperbola)
• \(x^2 + y^2 = 16\) (circle)

To resolve this type of system, the strategy often used involves substitution, elimination, or using matrices, depending on the complexity. Since these were nonlinear equations, the solution required rearranging one equation (circle) to make substitution into the other (hyperbola) possible. Through these methods, points \((0, 4)\) and \((0, -4)\) were found to be intersections, meaning that these points lie on both the hyperbola and the circle.