Problem 33 - Algebra 2

Question

SATELLITES For Exercises \(33-35,\) use the following information. Two satellites are placed in orbit about Earth. The equations of the two orbits \(\operatorname{are} \frac{x^{2}}{(300)^{2}}+\frac{y^{2}}{(900)^{2}}=1\) and \(\frac{x^{2}}{(600)^{2}}+\frac{y^{2}}{(690)^{2}}=1,\) where distances are in kilometers and Earth is the center of each curve. Solve each equation for \(y\)

Step-by-Step Solution

Answer

\(y = \pm\sqrt{810,000(1 - \frac{x^2}{90,000})}\) for the first orbit and \(y = \pm\sqrt{476,100(1 - \frac{x^2}{360,000})}\) for the second orbit.

Step 1: Identify the type of conic section

Both given equations are in the form \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), which indicates they are equations of ellipses with their centers at the origin \((0, 0)\).

Step 2: Solve the first equation for y

The first equation is \(\frac{x^{2}}{300^{2}}+\frac{y^{2}}{900^{2}}=1\).First, isolate \(y^2\):\[\frac{y^2}{900^2} = 1 - \frac{x^2}{300^2}.\] Multiply both sides by 810,000 (which is \(900^2\)) to solve for \(y^2\):\[y^2 = 810,000 \left(1 - \frac{x^2}{90,000}\right).\] Take the square root of both sides to solve for \(y\):\[y = \pm\sqrt{810,000 \left(1 - \frac{x^2}{90,000}\right)}.\]

Step 3: Solve the second equation for y

The second equation is \(\frac{x^{2}}{600^{2}}+\frac{y^{2}}{690^{2}}=1\).First, isolate \(y^2\):\[\frac{y^2}{690^2} = 1 - \frac{x^2}{600^2}.\] Multiply both sides by 476,100 (which is \(690^2\)) to solve for \(y^2\):\[y^2 = 476,100 \left(1 - \frac{x^2}{360,000}\right).\] Take the square root of both sides to solve for \(y\):\[y = \pm\sqrt{476,100 \left(1 - \frac{x^2}{360,000}\right)}.\]

Key Concepts

Conic Sections

Conic sections are created by the intersection of a plane and a double-napped cone. These sections result in four distinct shapes: circle, ellipse, parabola, and hyperbola. Each shape has unique properties and equations. An ellipse is formed when the plane cuts through one nappe of the cone at an angle, but doesn't intersect the base.
This gives rise to the general equation of an ellipse, which is in the form \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\). An ellipse is characterized by its two axes: 

• Major axis, which is the longest diameter.
• Minor axis, the shortest diameter.

This equation helps determine the shape and orientation, with the center located at the origin \((0, 0)\). All points on the ellipse are at a constant combined distance from two fixed points known as foci.
In this context, the exercise involves identifying elliptic equations and solving for the variable \(y\).

Orbit Equations

Orbit equations describe the paths of celestial bodies around a central focus, often an astronomical object like a planet or star. Satellites, for instance, follow elliptic orbits where Earth acts as one of the foci of the ellipse.
Each satellite's path can be modeled by an equation analogous to that of an ellipse, which may be centered at the origin when simplified. For precise calculations, one might need to determine the distances of the major and minor axes from the central point, indicating the extent and orientation of the orbit.
This allows us to predict satellite positions and understand the dynamics of their motion in space. For the case in our exercise, the expressions \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) clearly exemplify such orbit equations, where the values of \(a\) and \(b\) are derived from the maximum and minimum distances from the center.

Algebraic Manipulation

Algebraic manipulation involves applying mathematical transformations to simplify or rewrite expressions and equations. In the context of the exercise, algebraic manipulation is essential for isolating \(y^2\) in each orbit equation.
The process generally includes:

• Rearranging terms to isolate the desired variable on one side of the equation.
• Multiplying through by constants to eliminate fractions.
• Taking square roots as necessary to solve for variables directly.

This linear and systematic approach ensures that variables are clearly expressed, facilitating further computations or interpretations.
In solving these elliptic equations, careful manipulation helps to deduce expressions for \(y\), highlighting the dynamic nature of algebra in contributing to problem-solving.

Solving for Y

Solving for \(y\) in ellipse equations requires isolating \(y\), often involving steps like isolating \(y^2\), solving using algebraic manipulation, and finally, applying the square root to find \(y\). Start by identifying that the term \(\frac{y^2}{b^2}\) needs to be isolated on one side of the equation.
For example, rearrange the equation as \(\frac{y^2}{b^2} = 1 - \frac{x^2}{a^2}\). Then, multiply through by \(b^2\) to clear the fraction: \(y^2 = b^2 \left(1 - \frac{x^2}{a^2}\right)\).
The last step involves taking the square root on both sides, leading to \(y = \pm \sqrt{b^2 \left(1 - \frac{x^2}{a^2}\right)}\). The use of positive and negative signs indicates the two possible values of \(y\) for each \(x\), corresponding to the upper and lower halves of the ellipse. This method can be adapted to a variety of problems where similar transformations and calculations are needed.