Problem 63 - Precalculus: Mathematics for Calculus

Question

Determine what the value of \(F\) must be if the graph of the equation $$4 x^{2}+y^{2}+4(x-2 y)+F=0$$ is (a) an ellipse, (b) a single point, or (c) the empty set.

Step-by-Step Solution

Answer

(a) F < 5; (b) F = 5; (c) F > 5.

Step 1: Recognize the General Conic Form

The given equation is \(4x^2 + y^2 + 4(x - 2y) + F = 0\). Let's rewrite this in a standard conic form: \(Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0\). Here, \(A = 4\), \(B = 0\), \(C = 1\), \(D = 4\), \(E = -8\). We will analyze these coefficients to determine the type of conic section.

Step 2: Determine Conditions for an Ellipse

For the equation to represent an ellipse, the coefficients should satisfy \(B^2 - 4AC < 0\). Since \(B = 0\), the condition simplifies to \(-4AC < 0\). Therefore, \(-16 < 0\), which holds true, so the equation generally represents an ellipse. However, we need to complete the square to find the specific value of \(F\).

Step 3: Complete the Square for the Simplified Form

Start by rewriting the equation as \(4(x^2 + x) + (y^2 - 4y) = -F\). For the \(x\) terms, \(x^2 + x\) can be written as \((x + \frac{1}{2})^2 - \frac{1}{4}\). For the \(y\) terms, \(y^2 - 4y\) is \((y - 2)^2 - 4\). Substitute back: \[4((x+\frac{1}{2})^2 - \frac{1}{4}) + ((y-2)^2 - 4) = -F\] Which simplifies to: \[4(x+\frac{1}{2})^2 + (y - 2)^2 - 5 = -F\].

Step 4: Solve for Ellipse Condition

For the equation \(4(x+\frac{1}{2})^2 + (y-2)^2 = 5 - F\) to represent an ellipse, \(5 - F > 0\). Therefore, \(F < 5\). Thus, for an ellipse, \(F\) can be any value less than 5.

Step 5: Determine the Single Point Solution

The equation could represent a single point if \(4(x+\frac{1}{2})^2 + (y-2)^2 = 0\). This requires \(5 - F = 0\), solving gives \(F = 5\). Hence, the single point condition is \(F = 5\).

Step 6: Determine the Empty Set Condition

The equation represents the empty set if the terms on the left are positive, and \(5 - F < 0\). Therefore, \(F > 5\). Hence, for an empty set scenario, \(F > 5\).

Key Concepts

Ellipse

An ellipse is an elongated circle, appearing as an oval shape. It is defined as the set of all points such that the sum of the distances from two fixed points, called foci, is constant. When analyzing conic sections, an ellipse is identified through a specific condition on its equation coefficients.
Generally, an equation of the form \(Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0\) represents an ellipse if it satisfies\(B^2 - 4AC < 0\). In simpler terms, this inequality ensures that the equation does not describe other conic shapes like parabolas or hyperbolas.

• The coefficients \(A\), \(B\), and \(C\) determine the shape and orientation of the ellipse.
• Completing the square helps understand the specific form and position of the ellipse graph.

For the exercise given, by isolating \(F\) (the constant term) and ensuring that the inequality holds, we already know these coefficients form the basis for an ellipse for any \(F < 5\).

Single Point

A single point conic occurs when the whole equation's solutions boil down to a single specific location. In the context of conic sections, this means the geometric form reduces to a degenerate point. Here, each part of the equation must be satisfied by that one point, resulting in zeroes for each distance involved.
The equation \(4(x+rac{1}{2})^2 + (y-2)^2 = 0\) signifies a single point only if the right side is exactly zero.
The term \(5 - F = 0\) ensures that \(x\) and \(y\) can precisely balance each other out with their squared terms:

• Each squared term must individually equal zero.
• The point will be exactly the solution \(x = -\frac{1}{2}\) and \(y = 2\).

This precise solution can only exist if \(F = 5\), giving us the exact single point scenario.

Empty Set

The empty set, in this context, implies no real solution exists that can satisfy the equation. This occurs when the equation demands a negative right side, suggesting an impossible condition: a sum of positive squares equating to a negative number.
Since the squared terms \(4(x+rac{1}{2})^2 + (y-2)^2\) are always non-negative, the right side must also be non-negative to match. However, if \(5 - F < 0\), the balance breaks.

• This happens when \(F > 5\).
• The terms that are sums of squares can't possibly add to a negative number.

Thus, the equation becomes unsolvable within the real numbers, leading to the empty set condition.

Completing the Square

Completing the square is a handy mathematical technique used to simplify quadratic equations by turning them into a perfect square trinomial plus a constant. This is crucial when analyzing conic sections because it allows us to reformulate the equation into a recognizable geometric form.
This process involves:

• Taking a quadratic term, such as \(x^2 + x\) or \(y^2 - 4y\), and rewriting it as the square of a binomial.
• Adding and subtracting a specific number to make it a perfect square trinomial.

In our exercise, consider:
\(x^2 + x = (x + \frac{1}{2})^2 - \frac{1}{4}\), and \(y^2 - 4y = (y - 2)^2 - 4\).
The result simplifies the equation, revealing more about \(F\)'s impact and guiding conditions for ellipses, points, or empty sets.

Quadratic Equation Analysis

Quadratic equation analysis goes hand-in-hand with identifying conic sections. It involves scrutinizing the parabolic, hyperbolic, or elliptic nature of quadratic equations that include terms \(x^2\) and \(y^2\).
Key points include:

• Evaluating coefficients \(A\), \(B\), \(C\).
• Determining the nature using conditions such as \(B^2 - 4AC\).
• Recognizing when minimal or degenerate solutions occur, like points or empty sets.

In completing the square, we simplify and manipulate terms to understand these scenarios further. The analysis brings out geometric interpretations of real-world applications, helping clarify which values, like \(F\) here, satisfy specific conditions. Manipulating \(F\) offers a deeper insight into how the graph's form shifts or disappears altogether based on quadratic properties.