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

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

( x 2 ) 2 7 2 ( y + 7 ) 2 7 2 = 1 ; vertices: ( 9 , 7 ) , ( 5 , 7 ) ; foci: ( 2 + 7 2 , 7 ) , ( 2 7 2 , 7 ) ; asymptotes: y = x 9 , y = x 5

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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

( x + 3 ) 2 3 2 ( y 3 ) 2 3 2 = 1 ; vertices: ( 0 , 3 ) , ( 6 , 3 ) ; foci: ( 3 + 3 2 , 1 ) , ( 3 3 2 , 1 ) ; asymptotes: y = x + 6 , y = x

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

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4 x 2 + 24 x + 16 y 2 128 y + 156 = 0

( y 4 ) 2 2 2 ( x 3 ) 2 4 2 = 1 ; vertices: ( 3 , 6 ) , ( 3 , 2 ) ; foci: ( 3 , 4 + 2 5 ) , ( 3 , 4 2 5 ) ; asymptotes: y = 1 2 ( x 3 ) + 4 , y = 1 2 ( x 3 ) + 4

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

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

( y + 5 ) 2 7 2 ( x + 1 ) 2 70 2 = 1 ; vertices: ( 1 , 2 ) , ( 1 , 12 ) ; foci: ( 1 , 5 + 7 101 ) , ( 1 , 5 7 101 ) ; asymptotes: y = 1 10 ( x + 1 ) 5 , y = 1 10 ( x + 1 ) 5

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

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4 x 2 + 24 x 25 y 2 + 200 y 464 = 0

( x + 3 ) 2 5 2 ( y 4 ) 2 2 2 = 1 ; vertices: ( 2 , 4 ) , ( 8 , 4 ) ; foci: ( 3 + 29 , 4 ) , ( 3 29 , 4 ) ; asymptotes: y = 2 5 ( x + 3 ) + 4 , y = 2 5 ( x + 3 ) + 4

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For the following exercises, find the equations of the asymptotes for each hyperbola.

( x 3 ) 2 5 2 ( y + 4 ) 2 2 2 = 1

y = 2 5 ( x 3 ) 4 , y = 2 5 ( x 3 ) 4

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( y 3 ) 2 3 2 ( x + 5 ) 2 6 2 = 1

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9 x 2 18 x 16 y 2 + 32 y 151 = 0

y = 3 4 ( x 1 ) + 1 , y = 3 4 ( x 1 ) + 1

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16 y 2 + 96 y 4 x 2 + 16 x + 112 = 0

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Graphical

For the following exercises, sketch a graph of the hyperbola, labeling vertices and foci.

( y + 5 ) 2 9 ( x 4 ) 2 25 = 1

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( x 2 ) 2 8 ( y + 3 ) 2 27 = 1

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

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4 x 2 8 x + 16 y 2 32 y 52 = 0

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x 2 8 x 25 y 2 100 y 109 = 0

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x 2 + 8 x + 4 y 2 40 y + 88 = 0

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64 x 2 + 128 x 9 y 2 72 y 656 = 0

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

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100 x 2 + 1000 x + y 2 10 y 2575 = 0

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

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For the following exercises, given information about the graph of the hyperbola, find its equation.

Vertices at ( 3 , 0 ) and ( −3 , 0 ) and one focus at ( 5 , 0 ) .

x 2 9 y 2 16 = 1

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Vertices at ( 0 , 6 ) and ( 0 , −6 ) and one focus at ( 0 , −8 ) .

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Vertices at ( 1 , 1 ) and ( 11 , 1 ) and one focus at ( 12 , 1 ) .

( x 6 ) 2 25 ( y 1 ) 2 11 = 1

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Center: ( 0 , 0 ) ; vertex: ( 0 , −13 ) ; one focus: ( 0 , 313 ) .

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Center: ( 4 , 2 ) ; vertex: ( 9 , 2 ) ; one focus: ( 4 + 26 , 2 ) .

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

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Center: ( 3 , 5 ) ; vertex: ( 3 , 11 ) ; one focus: ( 3 , 5 + 2 10 ) .

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For the following exercises, given the graph of the hyperbola, find its equation.

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

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

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Extensions

For the following exercises, express the equation for the hyperbola as two functions, with y as a function of x . Express as simply as possible. Use a graphing calculator to sketch the graph of the two functions on the same axes.

y 2 9 x 2 1 = 1

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

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( x 2 ) 2 16 ( y + 3 ) 2 25 = 1

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4 x 2 16 x + y 2 2 y 19 = 0

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

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

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Real-world applications

For the following exercises, a hedge is to be constructed in the shape of a hyperbola near a fountain at the center of the yard. Find the equation of the hyperbola and sketch the graph.

The hedge will follow the asymptotes y = x  and  y = x , and its closest distance to the center fountain is 5 yards.

x 2 25 y 2 25 = 1

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The hedge will follow the asymptotes y = 2 x  and  y = −2 x , and its closest distance to the center fountain is 6 yards.

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The hedge will follow the asymptotes y = 1 2 x and y = 1 2 x , and its closest distance to the center fountain is 10 yards.

x 2 100 y 2 25 = 1

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The hedge will follow the asymptotes y = 2 3 x and y = 2 3 x , and its closest distance to the center fountain is 12 yards.

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The hedge will follow the asymptotes   y = 3 4 x  and  y = 3 4 x , and its closest distance to the center fountain is 20 yards.

x 2 400 y 2 225 = 1

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For the following exercises, assume an object enters our solar system and we want to graph its path on a coordinate system with the sun at the origin and the x-axis as the axis of symmetry for the object's path. Give the equation of the flight path of each object using the given information.

The object enters along a path approximated by the line y = x 2 and passes within 1 au (astronomical unit) of the sun at its closest approach, so that the sun is one focus of the hyperbola. It then departs the solar system along a path approximated by the line y = x + 2.

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The object enters along a path approximated by the line y = 2 x 2 and passes within 0.5 au of the sun at its closest approach, so the sun is one focus of the hyperbola. It then departs the solar system along a path approximated by the line y = −2 x + 2.

( x 1 ) 2 0.25 y 2 0.75 = 1

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The object enters along a path approximated by the line y = 0.5 x + 2 and passes within 1 au of the sun at its closest approach, so the sun is one focus of the hyperbola. It then departs the solar system along a path approximated by the line y = −0.5 x 2.

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The object enters along a path approximated by the line y = 1 3 x 1 and passes within 1 au of the sun at its closest approach, so the sun is one focus of the hyperbola. It then departs the solar system along a path approximated by the line   y = 1 3 x + 1.

( x 3 ) 2 4 y 2 5 = 1

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The object It enters along a path approximated by the line y = 3 x 9 and passes within 1 au of the sun at its closest approach, so the sun is one focus of the hyperbola. It then departs the solar system along a path approximated by the line y = −3 x + 9.

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Questions & Answers

if theta =30degree so COS2 theta = 1- 10 square theta upon 1 + tan squared theta
Martin Reply
how to compute this 1. g(1-x) 2. f(x-2) 3. g (-x-/5) 4. f (x)- g (x)
Yanah Reply
hi
John
hi
Grace
what sup friend
John
not much For functions, there are two conditions for a function to be the inverse function:   1--- g(f(x)) = x for all x in the domain of f     2---f(g(x)) = x for all x in the domain of g Notice in both cases you will get back to the  element that you started with, namely, x.
Grace
sin theta=3/4.prove that sec square theta barabar 1 + tan square theta by cosec square theta minus cos square theta
Umesh Reply
acha se dhek ke bata sin theta ke value
Ajay
sin theta ke ja gha sin square theta hoga
Ajay
I want to know trigonometry but I can't understand it anyone who can help
Siyabonga Reply
Yh
Idowu
which part of trig?
Nyemba
functions
Siyabonga
trigonometry
Ganapathi
differentiation doubhts
Ganapathi
hi
Ganapathi
hello
Brittany
Prove that 4sin50-3tan 50=1
Sudip Reply
f(x)= 1 x    f(x)=1x  is shifted down 4 units and to the right 3 units.
Sebit Reply
f (x) = −3x + 5 and g (x) = x − 5 /−3
Sebit
what are real numbers
Marty Reply
I want to know partial fraction Decomposition.
Adama Reply
classes of function in mathematics
Yazidu Reply
divide y2_8y2+5y2/y2
Sumanth Reply
wish i knew calculus to understand what's going on 🙂
Dashawn Reply
@dashawn ... in simple terms, a derivative is the tangent line of the function. which gives the rate of change at that instant. to calculate. given f(x)==ax^n. then f'(x)=n*ax^n-1 . hope that help.
Christopher
thanks bro
Dashawn
maybe when i start calculus in a few months i won't be that lost 😎
Dashawn
what's the derivative of 4x^6
Axmed Reply
24x^5
James
10x
Axmed
24X^5
Taieb
Thanks for this helpfull app
Axmed Reply
secA+tanA=2√5,sinA=?
richa Reply
tan2a+tan2a=√3
Rahulkumar
classes of function
Yazidu
if sinx°=sin@, then @ is - ?
NAVJIT Reply
the value of tan15°•tan20°•tan70°•tan75° -
NAVJIT
Practice Key Terms 4

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Source:  OpenStax, Algebra and trigonometry. OpenStax CNX. Nov 14, 2016 Download for free at https://legacy.cnx.org/content/col11758/1.6
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