Textbook Question
The equation of the red ellipse in the figure shown is x^2/25 + y^2/9 =1Write the equation for each circle shown in the figure.
1039
views
Ch. 7 - Conic Sections
Blitzer8th EditionCollege AlgebraISBN: 9780136970514
Not the one you use?Change textbookSelect a different textbook
Table of contents
- Ch. P - Fundamental Concepts of Algebra311
- Ch. 1 - Equations and Inequalities398
- Ch. 2 - Functions and Graphs407
- Ch. 3 - Polynomial and Rational Functions306
- Ch. 4 - Exponential and Logarithmic Functions247
- Ch. 5 - Systems of Equations and Inequalities173
- Ch. 6 - Matrices and Determinants143
- Ch. 7 - Conic Sections124
- Ch. 8 - Sequences, Induction, and Probability251
Chapter 8, Problem 67
Graph each semiellipse. y = -√16 - 4x²
Verified step by step guidance1
Recognize that the given equation is \(y = \sqrt{16 - 4x^{2}}\), which represents the upper half of an ellipse because \(y\) is defined as the positive square root.
Rewrite the equation by squaring both sides to eliminate the square root: \(y^{2} = 16 - 4x^{2}\).
Rearrange the equation to standard ellipse form by moving all terms to one side: \(4x^{2} + y^{2} = 16\).
Divide the entire equation by 16 to normalize it: \(\frac{4x^{2}}{16} + \frac{y^{2}}{16} = 1\), which simplifies to \(\frac{x^{2}}{4} + \frac{y^{2}}{16} = 1\).
Identify the ellipse parameters: the semi-major axis length \(a = 4\) (along the y-axis) and the semi-minor axis length \(b = 2\) (along the x-axis). Since \(y = \sqrt{16 - 4x^{2}}\) only gives the upper half, graph the ellipse only for \(y \geq 0\).
Verified video answer for a similar problem:
This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
5mWas this helpful?
Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Equation of an Ellipse
An ellipse is a set of points where the sum of distances to two foci is constant. Its standard form is (x²/a²) + (y²/b²) = 1. The given equation y = √(16 - 4x²) represents the upper half of an ellipse after rearranging to fit this form.
Recommended video:
5:30
Foci and Vertices of an Ellipse
Graphing Functions with Square Roots
When graphing y = √(expression), the output y is always non-negative, representing only the upper part of the curve. This means the graph is a semiellipse (half ellipse) above the x-axis, and the domain is restricted to values where the expression inside the root is non-negative.
Recommended video:
02:20Imaginary Roots with the Square Root Property
Domain and Range of the Function
The domain consists of all x-values for which the expression under the square root is ≥ 0. For y = √(16 - 4x²), this means 16 - 4x² ≥ 0, limiting x between -2 and 2. The range is y ≥ 0 since the square root function outputs non-negative values.
Recommended video:
4:22Domain & Range of Transformed Functions
Related Practice
Textbook Question
Find the solution set for each system by graphing both of the system's equations in the same rectangular coordinate system and finding points of intersection. Check all solutions in both equations.
924
views
Textbook Question
Find the standard form of the equation of an ellipse with vertices at (0, -6) and (0, 6), passing through (2, 4).
1346
views
Textbook Question
In Exercises 63–68, find the solution set for each system by graphing both of the system's equations in the same rectangular coordinate system and finding points of intersection. Check all solutions in both equations.
704
views
Textbook Question
Find the standard form of the equation of the hyperbola with vertices (5, −6) and (5, 6), passing through (0, 9).
1147
views
Textbook Question
Find the solution set for each system by graphing both of the system's equations in the same rectangular coordinate system and finding points of intersection. Check all solutions in both equations.
954
views