Problem 81 - Precalculus with Limits

Question

A circle and a parabola can have 0, 1, 2, 3, or 4 points of intersection. Sketch the circle given by \(x^2+y^2=4\). Discuss how this circle could intersect a parabola with an equation of the form \(y=x^2+C\). Then find the values of \(C\) for each of the five cases described below.Use a graphing utility to verify your results. (a) No points of intersection (b) One point of intersection (c) Two points of intersection (d) Three points of intersection (e) Four points of intersection

Step-by-Step Solution

Answer

Values of \(C\) for various intersections:(a) No intersections: \(C < -4\) or \(C > 4\)(b) One intersection: \(C = \pm 4\)(c) Two intersections: \(-4 < C < 4\)(d) Three Intersections: Not possible (e) Four Intersections: Not possible

Step 1: Sketch the circle

First sketch the circle \(x^2+y^2=4\) which is a circle of radius 2 centered at the origin. Draw this on an X-Y graph.

Step 2: Understand the Parabola

The general equation of the parabola given is \(y=x^2+C\), where C is a constant that adjusts the parabola vertically. Higher values of C would move the parabola upwards.

Step 3: Solve for C for no intersection

For no intersections, the parabola needs to lie entirely either above or below the circle. Therefore solve the inequality \(x^2 > 4\) or \(x^2 < -4\). Solving will give values of \(C < -4\) or \(C > 4\)

Step 4: Solve for C for one point intersection

For one point of intersection, the parabola must just touch the circle from above or below. Hence, it's either the vertex of the parabola at the top/bottom of circle. The equation is \(C = \pm 4\)

Step 5: Solve for C for two points of intersection

For two points of intersection, the parabola cuts the circle at two distinct points. The vertices of the parabola must be inside the circle. Solve for \( -4 < C < 4\)

Step 6: Solve for C for three points of intersection

For three points of intersection, the vertex of parabola must lie on circle and the parabola opens upwards. There are no real values of C to satisfy this condition as a parabola and a circle can intersect at most in 2 distinct real points.

Step 7: Solve for C for four points of intersection

Four intersections would appear only if there is some deformation either in the parabola or in the circle as a circle and parabola can't intersect at 4 points. No such C exist

Key Concepts

Circle-Parabola Intersection

Understanding the intersection points between a circle and a parabola can be fascinating. The circle equation given is \(x^2+y^2=4\), which represents a circle centered at the origin with a radius of 2.
On the other hand, we have a parabola described by \(y=x^2+C\), where \(C\) is a constant that shifts the parabola up or down. Depending on the value of \(C\), the parabola can intersect with the circle in different ways. Here's the interesting part:

• With no points of intersection, the parabola is entirely above or below the circle. This happens when \(C < -4\) or \(C > 4\).
• For one point of intersection, the vertex of the parabola just touches the circle, with values \(C = \pm 4\).
• Two points of intersection occur when the parabola cuts through the circle at distinct points, which requires \(-4 < C < 4\).
• Although theoretically Proposed, three or four intersections do not practically work with a standard circle and parabola combination.

These conditions organize the potential intersections based on varying parameter \(C\), demonstrating the dynamic relationship between a circle and a parabola.

Vertex of Parabola

The vertex of a parabola is a crucial feature in determining its intersection with other curves, like a circle. For the parabola \(y = x^2 + C\), the vertex lies at the point (0, \(C\)).
Knowing where the vertex is located helps us predict how the parabola intersects the circle \(x^2 + y^2 = 4\). Here's a simple way to think of it:

• If \(C\) is positive and large, the vertex is high above the circle, reducing intersection chances.
• If \(C\) is zero, the vertex sits at the origin, perfectly aligned at the center of the circle.
• As \(C\) becomes negative, the vertex moves downward below the circle.

This understanding is key, as the vertex’s position directly influences whether the parabola intersects the circle tangentially or at multiple points.

Vertical Shifts

Vertical shifts in the parabola are controlled by the constant \(C\) in the equation \(y = x^2 + C\). By tweaking \(C\), we translate the parabola vertically. Here's why this matters:

• If \(C\) becomes more positive, the entire parabola shifts upwards.
• If \(C\) is decreased, moving into negative values, the parabola slides down the coordinate plane.

Vertical shifts are essential when considering the intersection with the circle \(x^2+y^2=4\). For instance:

• Values of \(C > 4\) guarantee no intersection, as the parabola is beyond the circle’s top boundary.
• A negative \(C\) pushes the parabola below the circle’s bottom half, possibly giving zero points of intersection as well.

This simple vertical kinematic introduces a flexible way to explore a parabola's spatial relationship with a circle.

Graphing Utilities

Graphing utilities offer robust tools for visualizing and verifying the intersection points between a circle and a parabola. These software programs can graphically demonstrate how changes in \(C\) affect the parabola’s position relative to the circle. Here’s how graphing tools are typically used:

• Graphing calculators or software can quickly plot the circle and the parabola on the same axes.
• Adjust \(C\)’s value through sliders to observe instant vertical shifts in the parabola.
• They help in confirming the theoretical conclusions by showing live how many intersection points are feasible for different \(C\) values.

By taking advantage of graphing utilities, students reaffirm their understanding of the algebraic solutions and can visually appreciate the dynamic interaction in conic sections.