Chapter 10: Problem 12
Find two points on the graph of \(\frac{x^{2}}{16}+\frac{y^{2}}{4}=1\) by letting \(x=2\) and finding the corresponding values of \(y .\)
Short Answer
Expert verified
The points are (2, \(\sqrt{3}\)) and (2, -\(\sqrt{3}\)).
Step by step solution
01
Substitute x into the Equation
Start by substituting the given value of \( x = 2 \) into the equation of the ellipse. This means we use \( x = 2 \) in the equation \( \frac{x^{2}}{16}+\frac{y^{2}}{4}=1 \). It becomes: \( \frac{2^{2}}{16} + \frac{y^{2}}{4} = 1 \).
02
Simplify the Equation
Calculate \( \frac{2^{2}}{16} \). Since \( 2^{2} = 4 \), this becomes \( \frac{4}{16} \) or \( \frac{1}{4} \). So the equation is now \( \frac{1}{4} + \frac{y^{2}}{4} = 1 \).
03
Solve for y²
Rearrange the equation to solve for \( y^{2} \). Subtract \( \frac{1}{4} \) from both sides: \( \frac{y^{2}}{4} = 1 - \frac{1}{4} = \frac{3}{4} \).
04
Find y
Multiply both sides by 4 to isolate \( y^{2} \): \( y^{2} = 3 \). Take the square root of both sides to find \( y \): \( y = \pm \sqrt{3} \).
05
Determine the Points on the Curve
Since \( y \) can be \( \sqrt{3} \) or \( -\sqrt{3} \), the two points on the graph are \( (2, \sqrt{3}) \) and \( (2, -\sqrt{3}) \).
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Mastering the Art of Solving Equations
Solving equations is like finding the pieces to a puzzle. You want to make sure your equation is clear and you know what each part represents. In this exercise, our equation, \( \frac{x^{2}}{16} + \frac{y^{2}}{4} = 1 \), represents an ellipse. The key to solving equations like this involves patiently substituting and rearranging terms.
- Start by substituting known values into the equation. Here, we plug in \( x = 2 \) which transforms the equation into \( \frac{4}{16} + \frac{y^2}{4} = 1 \).
- Next, simplify wherever possible. Replace \( \frac{4}{16} \) with \( \frac{1}{4} \) to make calculations easier.
- Finally, isolate the desired variable—in this case, \( y \). By subtracting \( \frac{1}{4} \) from both sides, you get \( \frac{y^2}{4} = \frac{3}{4} \), and by solving for \( y^2 \), you find \( y = \pm \sqrt{3} \).
Graphing Ellipses: An Artistic Approach to Mathematics
Graphing an ellipse involves visualizing a stretched circle. To graph an ellipse from the equation \( \frac{x^{2}}{16} + \frac{y^{2}}{4} = 1 \), you need to identify the semi-major and semi-minor axes. These determine the shape and size of your ellipse.
- The semi-major axis lies along the axis with the larger denominator, in this case, the \( x \)-axis. The length is the square root of 16, which is 4.
- The semi-minor axis is along the \( y \)-axis with a length of \( \sqrt{4} \), which is 2.
- Sketch your x and y axes, marking the lengths of the semi-axes from the center (0, 0) to obtain 4 units along the x-axis and 2 units along the y-axis for the full width and height.
Understanding Coordinate Points on an Ellipse
Coordinate points let us pinpoint exact locations on the graph of an ellipse. They are the (\( x \), \( y \)) values that satisfy the equation of the ellipse. Let's see how these points are determined.
- Substitute a given \( x \) or \( y \) value into the ellipse equation to find the other variable.
- For example, substituting \( x = 2 \), we found \( y = \pm \sqrt{3} \). Hence, the points \((2, \sqrt{3})\) and \((2, -\sqrt{3})\) lie on the ellipse.
- Each point is carefully calculated to fit the ellipse equation, ensuring the graph correctly follows its path.