Chapter 12: Problem 25
Sketch the graph of each ellipse. $$ \frac{(x+1)^{2}}{16}+\frac{(y-2)^{2}}{25}=1 $$
Short Answer
Expert verified
The center is (-1, 2), vertices are (-1, 7) and (-1, -3), and co-vertices are (3, 2) and (-5, 2).
Step by step solution
01
Identify Center of the Ellipse
The given equation of the ellipse is \ \( \frac{(x+1)^{2}}{16} + \frac{(y-2)^{2}}{25} = 1 \). Compare this with the standard form \ \( \frac{(x-h)^{2}}{a^2} + \frac{(y-k)^{2}}{b^2} = 1 \). Here, \( h = -1 \) and \( k = 2 \), so the center of the ellipse is at \( (-1, 2) \).
02
Determine Lengths of Semi-Major and Semi-Minor Axes
Identify \( a^2 \) and \( b^2 \) from the given equation. Here, \( a^2 = 25 \) and \( b^2 = 16 \). Thus, \( a = 5 \) and \( b = 4 \). \( a \) is associated with the y-term, so the semi-major axis is vertical. \( b \) is associated with the x-term, so the semi-minor axis is horizontal.
03
Locate Vertices Along the Major Axis
Since the major axis is vertical, add and subtract \( a = 5 \) to the y-coordinate of the center to find the vertices. The vertices are at \( (-1, 2+5) = (-1, 7) \) and \( (-1, 2-5) = (-1, -3) \).
04
Locate Co-Vertices Along the Minor Axis
Since the minor axis is horizontal, add and subtract \( b = 4 \) to the x-coordinate of the center to find the co-vertices. The co-vertices are at \( (-1+4, 2) = (3, 2) \) and \( (-1-4, 2) = (-5, 2) \).
05
Sketch the Ellipse
Plot the center at \( (-1, 2) \) on the coordinate plane. Then plot the vertices at \( (-1, 7) \) and \( (-1, -3) \), and co-vertices at \( (3, 2) \) and \( (-5, 2) \). Draw a smooth, oval shape connecting these points to complete the graph of the ellipse.
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.
center of ellipse
The center of an ellipse is a crucial starting point in graphing it. For the given ellipse equation \(\frac{(x+1)^{2}}{16} + \frac{(y-2)^{2}}{25} = 1\), we compare it with the standard form \( \frac{(x-h)^{2}}{a^2} + \frac{(y-k)^{2}}{b^2} = 1 \). From this, we identify the center \( (h, k) \). Here, \( h = -1 \) and \( k = 2 \). Hence, the center of our ellipse is \( (-1, 2) \). This point is where both the major and minor axes intersect.
semi-major axis
The semi-major axis is the longest radius of the ellipse, stretching from the center to the furthest edge along the major axis. In our equation, \( a^2 = 25 \) indicates \( a = 5 \). This suggests a vertical semi-major axis since \( a \) is linked to the y-term. The length of the semi-major axis is \( 2a \), thus it spans 10 units in total – 5 units up and 5 units down from the center.
semi-minor axis
The semi-minor axis is the shortest radius of the ellipse, running from the center to the edge along the minor axis. From the equation, \( b^2 = 16 \) gives \( b = 4 \). The semi-minor axis is horizontal as \( b \) is associated with the x-term. The total length of the semi-minor axis is \( 2b \), totaling 8 units – 4 units to the right and 4 units to the left from the center.
vertices of ellipse
Vertices are the endpoints of the major axis. Since the major axis is vertical, we add and subtract \( a = 5 \) from the y-coordinate of the center. The vertices thus lie at \( (-1, 7) \) and \( (-1, -3) \). These points represent the highest and lowest points of the ellipse.
co-vertices
Co-vertices are the endpoints of the minor axis. Given our minor axis is horizontal, we add and subtract \( b = 4 \) to the x-coordinate of the center. Thus, co-vertices are at \( (3, 2) \) and \( (-5, 2) \). These points highlight the widest horizontal span of the ellipse. Plotting these along with the vertices gives a framework for sketching the ellipse's shape.