
Ellipse Review (11/15/24)
Flashcard
•
Mathematics
•
11th Grade
•
Practice Problem
•
Hard
Wayground Content
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15 questions
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1.
FLASHCARD QUESTION
Front
What is the standard form of the equation of an ellipse centered at the origin?
Back
The standard form is \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) where \( a \) is the semi-major axis and \( b \) is the semi-minor axis.
2.
FLASHCARD QUESTION
Front
What does it mean for an ellipse to be elongated horizontally?
Back
It means that the length of the semi-major axis (\( a \)) is greater than the length of the semi-minor axis (\( b \)), and the standard form of the equation is \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) with \( a^2 \) in the denominator of the \( x^2 \) term.
3.
FLASHCARD QUESTION
Front
What is the definition of an ellipse?
Back
An ellipse is the set of all points in a plane such that the sum of the distances from two fixed points (foci) is constant.
4.
FLASHCARD QUESTION
Front
How do you find the endpoints of the major axis of an ellipse?
Back
The endpoints of the major axis are located at (\( h \pm a \), \( k \)) for a horizontally oriented ellipse and (\( h \), \( k \pm a \)) for a vertically oriented ellipse, where (\( h, k \)) is the center of the ellipse.
5.
FLASHCARD QUESTION
Front
What are the coordinates of the foci of an ellipse?
Back
The foci are located at (\( h \pm c \), \( k \)) for a horizontally oriented ellipse and (\( h \), \( k \pm c \)) for a vertically oriented ellipse, where \( c = \sqrt{a^2 - b^2} \).
6.
FLASHCARD QUESTION
Front
What is the relationship between the semi-major axis, semi-minor axis, and the distance to the foci in an ellipse?
Back
In an ellipse, the relationship is given by \( c^2 = a^2 - b^2 \), where \( c \) is the distance from the center to each focus, \( a \) is the semi-major axis, and \( b \) is the semi-minor axis.
7.
FLASHCARD QUESTION
Front
What happens to the equation of an ellipse when it is translated?
Back
When an ellipse is translated, the center (h, k) changes, and the standard form of the equation becomes \( \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \).
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