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( x 2 ) 2 49 + ( y 4 ) 2 25 = 1

( x 2 ) 2 7 2 + ( y 4 ) 2 5 2 = 1 ; Endpoints of major axis ( 9 , 4 ) , ( 5 , 4 ) . Endpoints of minor axis ( 2 , 9 ) , ( 2 , 1 ) . Foci at ( 2 + 2 6 , 4 ) , ( 2 2 6 , 4 ) .

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( x 2 ) 2 81 + ( y + 1 ) 2 16 = 1

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( x + 5 ) 2 4 + ( y 7 ) 2 9 = 1

( x + 5 ) 2 2 2 + ( y 7 ) 2 3 2 = 1 ; Endpoints of major axis ( 5 , 10 ) , ( 5 , 4 ) . Endpoints of minor axis ( 3 , 7 ) , ( 7 , 7 ) . Foci at ( 5 , 7 + 5 ) , ( 5 , 7 5 ) .

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( x 7 ) 2 49 + ( y 7 ) 2 49 = 1

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4 x 2 8 x + 9 y 2 72 y + 112 = 0

( x 1 ) 2 3 2 + ( y 4 ) 2 2 2 = 1 ; Endpoints of major axis ( 4 , 4 ) , ( 2 , 4 ) . Endpoints of minor axis ( 1 , 6 ) , ( 1 , 2 ) . Foci at ( 1 + 5 , 4 ) , ( 1 5 , 4 ) .

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9 x 2 54 x + 9 y 2 54 y + 81 = 0

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4 x 2 24 x + 36 y 2 360 y + 864 = 0

( x 3 ) 2 ( 3 2 ) 2 + ( y 5 ) 2 ( 2 ) 2 = 1 ; Endpoints of major axis ( 3 + 3 2 , 5 ) , ( 3 3 2 , 5 ) . Endpoints of minor axis ( 3 , 5 + 2 ) , ( 3 , 5 2 ) . Foci at ( 7 , 5 ) , ( 1 , 5 ) .

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4 x 2 + 24 x + 16 y 2 128 y + 228 = 0

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4 x 2 + 40 x + 25 y 2 100 y + 100 = 0

( x + 5 ) 2 ( 5 ) 2 + ( y 2 ) 2 ( 2 ) 2 = 1 ; Endpoints of major axis ( 0 , 2 ) , ( 10 , 2 ) . Endpoints of minor axis ( 5 , 4 ) , ( 5 , 0 ) . Foci at ( 5 + 21 , 2 ) , ( 5 21 , 2 ) .

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x 2 + 2 x + 100 y 2 1000 y + 2401 = 0

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4 x 2 + 24 x + 25 y 2 + 200 y + 336 = 0

( x + 3 ) 2 ( 5 ) 2 + ( y + 4 ) 2 ( 2 ) 2 = 1 ; Endpoints of major axis ( 2 , 4 ) , ( 8 , 4 ) . Endpoints of minor axis ( 3 , 2 ) , ( 3 , 6 ) . Foci at ( 3 + 21 , 4 ) , ( 3 21 , 4 ) .

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9 x 2 + 72 x + 16 y 2 + 16 y + 4 = 0

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For the following exercises, find the foci for the given ellipses.

( x + 3 ) 2 25 + ( y + 1 ) 2 36 = 1

Foci ( 3 , 1 + 11 ) , ( 3 , 1 11 )

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( x + 1 ) 2 100 + ( y 2 ) 2 4 = 1

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x 2 + y 2 = 1

Focus ( 0 , 0 )

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x 2 + 4 y 2 + 4 x + 8 y = 1

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10 x 2 + y 2 + 200 x = 0

Foci ( 10 , 30 ) , ( 10 , 30 )

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Graphical

For the following exercises, graph the given ellipses, noting center, vertices, and foci.

x 2 16 + y 2 9 = 1

Center ( 0 , 0 ) , Vertices ( 4 , 0 ) , ( 4 , 0 ) , ( 0 , 3 ) , ( 0 , 3 ) , Foci ( 7 , 0 ) , ( 7 , 0 )

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81 x 2 + 49 y 2 = 1

Center ( 0 , 0 ) , Vertices ( 1 9 , 0 ) , ( 1 9 , 0 ) , ( 0 , 1 7 ) , ( 0 , 1 7 ) , Foci ( 0 , 4 2 63 ) , ( 0 , 4 2 63 )

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( x 2 ) 2 64 + ( y 4 ) 2 16 = 1

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( x + 3 ) 2 9 + ( y 3 ) 2 9 = 1

Center ( 3 , 3 ) , Vertices ( 0 , 3 ) , ( 6 , 3 ) , ( 3 , 0 ) , ( 3 , 6 ) , Focus ( 3 , 3 )

Note that this ellipse is a circle. The circle has only one focus, which coincides with the center.

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x 2 2 + ( y + 1 ) 2 5 = 1

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4 x 2 8 x + 16 y 2 32 y 44 = 0

Center ( 1 , 1 ) , Vertices ( 5 , 1 ) , ( 3 , 1 ) , ( 1 , 3 ) , ( 1 , 1 ) , Foci ( 1 , 1 + 4 3 ) , ( 1 , 1 4 3 )

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x 2 8 x + 25 y 2 100 y + 91 = 0

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x 2 + 8 x + 4 y 2 40 y + 112 = 0

Center ( 4 , 5 ) , Vertices ( 2 , 5 ) , ( 6 , 4 ) , ( 4 , 6 ) , ( 4 , 4 ) , Foci ( 4 + 3 , 5 ) , ( 4 3 , 5 )

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64 x 2 + 128 x + 9 y 2 72 y 368 = 0

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16 x 2 + 64 x + 4 y 2 8 y + 4 = 0

Center ( 2 , 1 ) , Vertices ( 0 , 1 ) , ( 4 , 1 ) , ( 2 , 5 ) , ( 2 , 3 ) , Foci ( 2 , 1 + 2 3 ) , ( 2 , 1 2 3 )

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100 x 2 + 1000 x + y 2 10 y + 2425 = 0

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4 x 2 + 16 x + 4 y 2 + 16 y + 16 = 0

Center ( 2 , 2 ) , Vertices ( 0 , 2 ) , ( 4 , 2 ) , ( 2 , 0 ) , ( 2 , 4 ) , Focus ( 2 , 2 )

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For the following exercises, use the given information about the graph of each ellipse to determine its equation.

Center at the origin, symmetric with respect to the x - and y -axes, focus at ( 4 , 0 ) , and point on graph ( 0 , 3 ) .

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Center at the origin, symmetric with respect to the x - and y -axes, focus at ( 0 , −2 ) , and point on graph ( 5 , 0 ) .

x 2 25 + y 2 29 = 1

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Center at the origin, symmetric with respect to the x - and y -axes, focus at ( 3 , 0 ) , and major axis is twice as long as minor axis.

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Center ( 4 , 2 ) ; vertex ( 9 , 2 ) ; one focus: ( 4 + 2 6 , 2 ) .

( x 4 ) 2 25 + ( y 2 ) 2 1 = 1

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Center ( 3 , 5 ) ; vertex ( 3 , 11 ) ; one focus: ( 3 ,  5+4 2 )

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Center ( −3 , 4 ) ; vertex ( 1 , 4 ) ; one focus: ( −3 + 2 3 , 4 )

( x + 3 ) 2 16 + ( y 4 ) 2 4 = 1

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For the following exercises, given the graph of the ellipse, determine its equation.

( x + 2 ) 2 4 + ( y 2 ) 2 9 = 1

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Extensions

For the following exercises, find the area of the ellipse. The area of an ellipse is given by the formula Area = a b π .

( x 3 ) 2 9 + ( y 3 ) 2 16 = 1

Area = 12π square units

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( x + 6 ) 2 16 + ( y 6 ) 2 36 = 1

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( x + 1 ) 2 4 + ( y 2 ) 2 5 = 1

Area = 2 5 π square units

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4 x 2 8 x + 9 y 2 72 y + 112 = 0

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9 x 2 54 x + 9 y 2 54 y + 81 = 0

Area = 9π square units

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Real-world applications

Find the equation of the ellipse that will just fit inside a box that is 8 units wide and 4 units high.

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Find the equation of the ellipse that will just fit inside a box that is four times as wide as it is high. Express in terms of h , the height.

x 2 4 h 2 + y 2 1 4 h 2 = 1

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An arch has the shape of a semi-ellipse (the top half of an ellipse). The arch has a height of 8 feet and a span of 20 feet. Find an equation for the ellipse, and use that to find the height to the nearest 0.01 foot of the arch at a distance of 4 feet from the center.

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An arch has the shape of a semi-ellipse. The arch has a height of 12 feet and a span of 40 feet. Find an equation for the ellipse, and use that to find the distance from the center to a point at which the height is 6 feet. Round to the nearest hundredth.

x 2 400 + y 2 144 = 1 . Distance = 17.32 feet

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A bridge is to be built in the shape of a semi-elliptical arch and is to have a span of 120 feet. The height of the arch at a distance of 40 feet from the center is to be 8 feet. Find the height of the arch at its center.

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A person in a whispering gallery standing at one focus of the ellipse can whisper and be heard by a person standing at the other focus because all the sound waves that reach the ceiling are reflected to the other person. If a whispering gallery has a length of 120 feet, and the foci are located 30 feet from the center, find the height of the ceiling at the center.

Approximately 51.96 feet

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A person is standing 8 feet from the nearest wall in a whispering gallery. If that person is at one focus, and the other focus is 80 feet away, what is the length and height at the center of the gallery?

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Questions & Answers

for the "hiking" mix, there are 1,000 pieces in the mix, containing 390.8 g of fat, and 165 g of protein. if there is the same amount of almonds as cashews, how many of each item is in the trail mix?
ADNAN Reply
linear speed of an object
Melissa Reply
an object is traveling around a circle with a radius of 13 meters .if in 20 seconds a central angle of 1/7 Radian is swept out what are the linear and angular speed of the object
Melissa
test
Matrix
how to find domain
Mohamed Reply
like this: (2)/(2-x) the aim is to see what will not be compatible with this rational expression. If x= 0 then the fraction is undefined since we cannot divide by zero. Therefore, the domain consist of all real numbers except 2.
Dan
define the term of domain
Moha
if a>0 then the graph is concave
Angel Reply
if a<0 then the graph is concave blank
Angel
what's a domain
Kamogelo Reply
The set of all values you can use as input into a function su h that the output each time will be defined, meaningful and real.
Spiro
how fast can i understand functions without much difficulty
Joe Reply
what is inequalities
Nathaniel
functions can be understood without a lot of difficulty. Observe the following: f(2) 2x - x 2(2)-2= 2 now observe this: (2,f(2)) ( 2, -2) 2(-x)+2 = -2 -4+2=-2
Dan
what is set?
Kelvin Reply
a colony of bacteria is growing exponentially doubling in size every 100 minutes. how much minutes will it take for the colony of bacteria to triple in size
Divya Reply
I got 300 minutes. is it right?
Patience
no. should be about 150 minutes.
Jason
It should be 158.5 minutes.
Mr
ok, thanks
Patience
100•3=300 300=50•2^x 6=2^x x=log_2(6) =2.5849625 so, 300=50•2^2.5849625 and, so, the # of bacteria will double every (100•2.5849625) = 258.49625 minutes
Thomas
158.5 This number can be developed by using algebra and logarithms. Begin by moving log(2) to the right hand side of the equation like this: t/100 log(2)= log(3) step 1: divide each side by log(2) t/100=1.58496250072 step 2: multiply each side by 100 to isolate t. t=158.49
Dan
what is the importance knowing the graph of circular functions?
Arabella Reply
can get some help basic precalculus
ismail Reply
What do you need help with?
Andrew
how to convert general to standard form with not perfect trinomial
Camalia Reply
can get some help inverse function
ismail
Rectangle coordinate
Asma Reply
how to find for x
Jhon Reply
it depends on the equation
Robert
yeah, it does. why do we attempt to gain all of them one side or the other?
Melissa
how to find x: 12x = 144 notice how 12 is being multiplied by x. Therefore division is needed to isolate x and whatever we do to one side of the equation we must do to the other. That develops this: x= 144/12 divide 144 by 12 to get x. addition: 12+x= 14 subtract 12 by each side. x =2
Dan
whats a domain
mike Reply
The domain of a function is the set of all input on which the function is defined. For example all real numbers are the Domain of any Polynomial function.
Spiro
Spiro; thanks for putting it out there like that, 😁
Melissa
foci (–7,–17) and (–7,17), the absolute value of the differenceof the distances of any point from the foci is 24.
Churlene Reply
Practice Key Terms 7

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Source:  OpenStax, Precalculus. OpenStax CNX. Jan 19, 2016 Download for free at https://legacy.cnx.org/content/col11667/1.6
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