Problem 25 - The Calculus with Analytic Geometry

Question

In Exercises 7 through 28 , draw a sketch of the graph of the equation. $$ 4 x^{2}+y^{2}=0 $$

Step-by-Step Solution

Answer

The given equation 4x^2 + y^2 = 0 has no real solutions, so there is no graph to draw.

Step 1: Identify the type of equation

Determine the type of equation given. The equation is written as 4x^2 + y^2 = 0.

Step 2: Analyze the equation

Consider the general forms of conic sections (ellipse, parabola, hyperbola) and compare. This equation resembles an ellipse but note that the right side equals zero.

Step 3: Solve for y in terms of x

Start by isolating one term, for example, y^2 = -4x^2. Given the terms, solve for y: y = sqrt(-4x^2).

Step 4: Evaluate the possible solutions

Evaluate the equation under realistic conditions. The term -4x^2 under the square root is an issue because it is negative. Therefore, no real solutions exist for y .

Step 5: Conclude the graph properties

Since there are no real solutions to y^2 = -4x^2, the graph doesn't represent any real points on the coordinate plane.

Key Concepts

Conic Sections

Conic sections are curves obtained from the intersection of a plane with a cone. These curves include ellipses, parabolas, hyperbolas, and circles. Each type of conic section has a specific general equation:

• Ellipse: \(Ax^2 + By^2 = C\) where A and B have the same sign and are positive
• Parabola: \(Ay = X^2 \) or \(Ax = Y^2 \)
• Hyperbola: \(Ax^2 - By^2 = C\) where A and B have opposite signs
• Circle: A special type of ellipse where A and B are equal

The equation 4x^2 + y^2 = 0 initially appears like an ellipse. However, analyzing further will reveal important details.

Ellipse Equation

An ellipse has the general form \(Ax^2 + By^2 = C\), where A, B, and C are constants, and A and B are positive but not necessarily equal.
For example, \(4x^2 + y^2 = 1\) forms an ellipse.
The given equation \(4x^2 + y^2 = 0\) also initially resembles an ellipse. The next steps include analyzing the equation to confirm its properties.
Let's rewrite the equation to isolate terms: \(y^2 = -4x^2\). We would then solve for y: \(\[\begin{equation}y = \sqrt{-4x^2}\end{equation}\]\). This leads to a critical point about real number solutions, which will be discussed next.

No Real Solutions

When solving \(y^2 = -4x^2\), consider the square root term. \(-4x^2\) is negative, and the square root of a negative number isn't a real number.
Real numbers do not include imaginary numbers, which are numbers involving \(\text{i}\), representing \(\text{the square root of -1}\). Hence, the equation \(\[\begin{equation}y = \sqrt{-4x^2}\end{equation}\]\) has no real solutions.
This indicates there are no points (x, y) that satisfy the equation \((4x^2 + y^2 = 0)\). Therefore, it doesn't graph on the coordinate plane as a real conic section. Understanding why no real solutions exist is key in evaluating and graphing the equation accurately.