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Chapter 1: Problem 25

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

Short Answer

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

Step by step solution

01

Identify the type of equation

Determine the type of equation given. The equation is written as 4x^2 + y^2 = 0.
02

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.
03

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).
04

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 .
05

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

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