Chapter 7: Problem 32
Translate the ellipse with the given equation so that it is centered at the given point. Find the new equation and sketch its graph. $$ \frac{x^{2}}{15}+\frac{y^{2}}{16}=1 ;(5,-6) $$
Short Answer
Expert verified
The new equation is \( \frac{(x - 5)^{2}}{15} + \frac{(y + 6)^{2}}{16} = 1 \).
Step by step solution
01
Understanding the Standard Form of the Ellipse
The given equation \(\frac{x^{2}}{15} + \frac{y^{2}}{16} = 1\) represents an ellipse centered at the origin \((0,0)\) with axes along the x and y directions. The term \(\frac{x^{2}}{15}\) indicates that the semi-major or semi-minor axis along the x-direction is \(\sqrt{15}\), and \(\frac{y^{2}}{16}\) indicates that the semi-major or semi-minor axis along the y-direction is 4.
02
Determine the Translation of the Center
To translate the center of the ellipse to point \((5, -6)\), the variables \(x\) and \(y\) need to be replaced in the equation. Translations affect the center by adjusting \(x\) to \(x - h\) and \(y\) to \(y - k\), where \((h, k)\) is the new center \((5, -6)\).
03
Substitute the Translated Variables into the Equation
Replace \(x\) with \((x - 5)\) and \(y\) with \((y + 6)\) in the original ellipse equation: \[ \frac{(x - 5)^{2}}{15} + \frac{(y + 6)^{2}}{16} = 1 \].
04
Simplified New Equation of the Translated Ellipse
After substituting, the new equation of the ellipse becomes:\[ \frac{(x - 5)^{2}}{15} + \frac{(y + 6)^{2}}{16} = 1 \]. This equation reflects the ellipse centered at \((5, -6)\) with unchanged semi-axis lengths.
05
Sketching the Graph of the Translated Ellipse
To sketch the graph, plot the center of the ellipse at \((5, -6)\) on a coordinate grid, then draw the ellipse using the semi-axis lengths: \(\sqrt{15}\) along the x-direction and 4 along the y-direction. The figure will maintain the orientation and proportions but shifted to the new center.
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.
Understanding the Ellipse Equation
An ellipse is a type of conic section that looks like an elongated circle. The standard form of an ellipse's equation on a 2D coordinate plane is:
The given ellipse equation \(\frac{x^{2}}{15} + \frac{y^{2}}{16} = 1\) suggests a vertical major axis since the denominator under \(y^2\) is larger. The ellipse is centered at \((0,0)\) and can be translated by modifying the center coordinates.
- Horizontal major axis: \( \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \)
- Vertical major axis: \( \frac{(x-h)^2}{b^2} + \frac{(y-k)^2}{a^2} = 1 \)
- \((h, k)\) are the coordinates of the center of the ellipse,
- \(a\) is the length of the semi-major axis,
- \(b\) is the length of the semi-minor axis.
The given ellipse equation \(\frac{x^{2}}{15} + \frac{y^{2}}{16} = 1\) suggests a vertical major axis since the denominator under \(y^2\) is larger. The ellipse is centered at \((0,0)\) and can be translated by modifying the center coordinates.
Coordinate Geometry in Ellipse Translation
Coordinate geometry helps us place and manipulate geometric figures in a plane. When you translate an ellipse, you're moving its center from one point to another without altering its shape or orientation.
In this exercise, the ellipse given by the equation \(\frac{x^{2}}{15} + \frac{y^{2}}{16} = 1\) is originally centered at the origin \((0,0)\). To translate it to the point \((5,-6)\), every x-coordinate of the ellipse is shifted 5 units right, and every y-coordinate is shifted 6 units down.This transformation can be done algebraically by replacing each \(x\) and \(y\) in the original equation with \((x-5)\) and \((y+6)\) respectively. This results in the translated equation:\[ \frac{(x-5)^{2}}{15} + \frac{(y+6)^{2}}{16} = 1 \]The transformation maintains the length of axes \(\sqrt{15}\) and \(4\), ensuring the ellipse retains its proportions and size post-translation.
In this exercise, the ellipse given by the equation \(\frac{x^{2}}{15} + \frac{y^{2}}{16} = 1\) is originally centered at the origin \((0,0)\). To translate it to the point \((5,-6)\), every x-coordinate of the ellipse is shifted 5 units right, and every y-coordinate is shifted 6 units down.This transformation can be done algebraically by replacing each \(x\) and \(y\) in the original equation with \((x-5)\) and \((y+6)\) respectively. This results in the translated equation:\[ \frac{(x-5)^{2}}{15} + \frac{(y+6)^{2}}{16} = 1 \]The transformation maintains the length of axes \(\sqrt{15}\) and \(4\), ensuring the ellipse retains its proportions and size post-translation.
Ellipses as Conic Sections
An ellipse is one of the four types of conic sections, which also include circles, parabolas, and hyperbolas. Conic sections are curves obtained by slicing a cone with a plane at different inclinations.
For ellipses, the plane cuts through both nappes of the cone at an angle that is not perpendicular to the cone's base. This creates a closed, symmetric curve.
In the study of conic sections, the ellipse is often the focus due to its unique properties:
In the study of conic sections, the ellipse is often the focus due to its unique properties:
- It has two focal points and the sum of the distances from any point on the ellipse to these foci is constant.
- Its reflective property is important in optics and physics, where it guides practical applications like designing elliptical mirrors or lenses.