Problem 47 - College Algebra

Question

A planet's orbit follows a path described by \(16 x^{2}+4 y^{2}=64 .\) A comet follows the parabolic path \(y=x^{2}-4 .\) Where might the comet intersect the orbiting planet?

Step-by-Step Solution

Answer

The comet might intersect the planet's orbit at points (0, -4), (-2, 0), and (2, 0)

Step 1: Expression Rewriting

We first rewrite the ellipse equation \(16 x^{2}+4 y^{2}=64 \) in a simplified form by dividing by 4 to get \(4x^2 + y^2 = 16\). This simplification makes the calculation process easier.

Step 2: Replacing y

Substitute \(y = x^2 - 4\) (the equation of the parabola) into the ellipse equation. This gives: \(4x^2 + (x^2 - 4)^2 = 16\). This simplifies to \(4x^2 + x^4 - 8x^2 + 16 = 16\).

Step 3: Simplifying to Quadratic Form

We simplify the equation derived from the earlier step to put it in quadratic format. The equation simplifies to \(x^4 - 4x^2 = 0\).

Step 4: Solving Quadratic Equation

Solve the equation \(x^4 - 4x^2 = 0\), by factoring to get \(x^2(x^2 - 4) = 0\). The solutions to this equation are \(x = 0, -2, 2\).

Step 5: Find Corresponding Y Values

Finally, we substitute each x value into the parabola equation to get the corresponding y values. When \(x = 0\), \(y = (0)^2 - 4 = -4\). When \(x = -2\) and \(2\), \(y = (-2)^2 - 4 = 0\) and \(y = (2)^2 - 4 = 0\), respectively. Thus, the intersection points are (0, -4) and (-2, 0), (2, 0).

Key Concepts

Ellipse

An ellipse is a geometric shape that resembles a flattened circle. In algebra, an ellipse is typically defined using an equation involving two variables, generally \(x\) and \(y\). The general standard form of an ellipse's equation is \((A)x^2 + (B)y^2 = (C)\), where \(A\), \(B\), and \(C\) are constants. The ellipse from our exercise is described by the equation \(4x^2 + y^2 = 16\).

• Here, the ellipse is centered at the origin (0,0) because there are no additional terms for \x\ and \(y\) other than their squares.
• The equation shows that the ellipse's axes are aligned with the Cartesian axes since \x\ and \y\ do not have \(x-y\) cross-product terms.
• To simplify and solve for the ellipse in context with another equation, you often transform its equation into a more manageable form, as done through division in the exercise.

The major and minor axes of an ellipse are determined by the coefficients of the squared terms. In the given equation \(4x^2 + y^2 = 16\), the semimajor and semiminor axes would traditionally require dividing the equation in a form relating \(x\) and \(y\) directly to these dimensions.

Parabola

A parabola represents the path of a quadratic equation and is a curve with a distinct U-shape. Every parabola has a vertex, which is its highest or lowest point, and an axis of symmetry, which is a line that divides the parabola into two mirror-image halves. The equation for a parabola is typically written as \(y = ax^2 + bx + c\), where \(a\), \(b\), and \(c\) are constants.In the exercise, the equation of the parabolic path is \(y = x^2 - 4\). This equation can directly help us find intersection points with other curves by substitution.

• The vertex of this parabola is located at (0, -4). This can be observed because the \(x^2\) term doesn't shift horizontally, and the -4 translates the parabola downward along the y-axis.
• The parabola is symmetric about the y-axis since there is no \(x\) term.
• In a solutions context, substituting this parabola's equation into another curve's equation transforms the geometry into solvable algebraic forms, as demonstrated in the step-by-step solution.

Using parabolas in algebra often involves factoring quadratic equations to pinpoint crucial points like intersections or vertices.

Intersection Points

Intersection points occur where two curves meet in a coordinate system. These points are crucial for determining geometric relationships and often reveal key insights in problem-solving. In the context of algebra, finding intersection points involves solving equations simultaneously since these points satisfy both equations of the involved curves.

• In our exercise, the intersection points between the ellipse \(4x^2 + y^2 = 16\) and the parabola \(y = x^2 - 4\) are discovered by substituting the expression for \(y\) from the parabola into the ellipse's equation.
• This approach reduces the two equations into a single expression with one variable, simplifying the process of finding common solutions or roots.
• The calculation shows intersections at three points: (0, -4), (-2, 0), and (2, 0). These points lie on both the parabola and the ellipse, and confirm the exact locations of intersection by substitution and solving methods.

Examining intersection points not only solves mathematical exercises but also translates to understanding overlapping paths or collided trajectories in physical space.

Quadratic Equations

Quadratic equations are fundamental in algebra and take the general form: \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are coefficients. Quadratics are key to solving problems involving curved paths such as parabolas.In the exercise, we start with the equation \(x^4 - 4x^2 = 0\), derived from substituting and rearranging the ellipse and parabola equations. This needs to be reduced to a quadratic form to be solvable.

• Factorization simplifies the quadratic equation, turning it into easily solvable parts. In the example, \(x^2(x^2 - 4) = 0\) simplifies further, isolating potential solutions of \(x^2 - 4 = 0\) and \(x^2 = 0\).
• Solving these equations, we find solutions for \(x\) as \x = 0, -2, \text{and}\ 2\.
• Each solution for \(x\) can then be plugged back into another equation (the parabola) to find corresponding \(y\)-values, illustrating how quadratic solutions tie into broader algebraic systems.

Mastering quadratic equations helps unlock more complex algebraic intersections and succession, illustrating how concepts transform and solve real-world problems in mathematical terms.