# 10.2 The hyperbola  (Page 9/13)

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$\frac{{\left(y-6\right)}^{2}}{36}-\frac{{\left(x+1\right)}^{2}}{16}=1$

$\frac{{\left(x-2\right)}^{2}}{49}-\frac{{\left(y+7\right)}^{2}}{49}=1$

$\frac{{\left(x-2\right)}^{2}}{{7}^{2}}-\frac{{\left(y+7\right)}^{2}}{{7}^{2}}=1;\text{\hspace{0.17em}}$ vertices: $\text{\hspace{0.17em}}\left(9,-7\right),\left(-5,-7\right);\text{\hspace{0.17em}}$ foci: $\text{\hspace{0.17em}}\left(2+7\sqrt{2},-7\right),\left(2-7\sqrt{2},-7\right);\text{\hspace{0.17em}}$ asymptotes: $\text{\hspace{0.17em}}y=x-9,y=-x-5$

$4{x}^{2}-8x-9{y}^{2}-72y+112=0$

$-9{x}^{2}-54x+9{y}^{2}-54y+81=0$

$\frac{{\left(x+3\right)}^{2}}{{3}^{2}}-\frac{{\left(y-3\right)}^{2}}{{3}^{2}}=1;\text{\hspace{0.17em}}$ vertices: $\text{\hspace{0.17em}}\left(0,3\right),\left(-6,3\right);\text{\hspace{0.17em}}$ foci: $\text{\hspace{0.17em}}\left(-3+3\sqrt{2},1\right),\left(-3-3\sqrt{2},1\right);\text{\hspace{0.17em}}$ asymptotes: $\text{\hspace{0.17em}}y=x+6,y=-x$

$4{x}^{2}-24x-36{y}^{2}-360y+864=0$

$-4{x}^{2}+24x+16{y}^{2}-128y+156=0$

$\frac{{\left(y-4\right)}^{2}}{{2}^{2}}-\frac{{\left(x-3\right)}^{2}}{{4}^{2}}=1;\text{\hspace{0.17em}}$ vertices: $\text{\hspace{0.17em}}\left(3,6\right),\left(3,2\right);\text{\hspace{0.17em}}$ foci: $\text{\hspace{0.17em}}\left(3,4+2\sqrt{5}\right),\left(3,4-2\sqrt{5}\right);\text{\hspace{0.17em}}$ asymptotes: $\text{\hspace{0.17em}}y=\frac{1}{2}\left(x-3\right)+4,y=-\frac{1}{2}\left(x-3\right)+4$

$-4{x}^{2}+40x+25{y}^{2}-100y+100=0$

${x}^{2}+2x-100{y}^{2}-1000y+2401=0$

$\frac{{\left(y+5\right)}^{2}}{{7}^{2}}-\frac{{\left(x+1\right)}^{2}}{{70}^{2}}=1;\text{\hspace{0.17em}}$ vertices: $\text{\hspace{0.17em}}\left(-1,2\right),\left(-1,-12\right);\text{\hspace{0.17em}}$ foci: $\text{\hspace{0.17em}}\left(-1,-5+7\sqrt{101}\right),\left(-1,-5-7\sqrt{101}\right);\text{\hspace{0.17em}}$ asymptotes: $\text{\hspace{0.17em}}y=\frac{1}{10}\left(x+1\right)-5,y=-\frac{1}{10}\left(x+1\right)-5$

$-9{x}^{2}+72x+16{y}^{2}+16y+4=0$

$4{x}^{2}+24x-25{y}^{2}+200y-464=0$

$\frac{{\left(x+3\right)}^{2}}{{5}^{2}}-\frac{{\left(y-4\right)}^{2}}{{2}^{2}}=1;\text{\hspace{0.17em}}$ vertices: $\text{\hspace{0.17em}}\left(2,4\right),\left(-8,4\right);\text{\hspace{0.17em}}$ foci: $\text{\hspace{0.17em}}\left(-3+\sqrt{29},4\right),\left(-3-\sqrt{29},4\right);\text{\hspace{0.17em}}$ asymptotes: $\text{\hspace{0.17em}}y=\frac{2}{5}\left(x+3\right)+4,y=-\frac{2}{5}\left(x+3\right)+4$

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

$\frac{{y}^{2}}{{3}^{2}}-\frac{{x}^{2}}{{3}^{2}}=1$

$\frac{{\left(x-3\right)}^{2}}{{5}^{2}}-\frac{{\left(y+4\right)}^{2}}{{2}^{2}}=1$

$y=\frac{2}{5}\left(x-3\right)-4,y=-\frac{2}{5}\left(x-3\right)-4$

$\frac{{\left(y-3\right)}^{2}}{{3}^{2}}-\frac{{\left(x+5\right)}^{2}}{{6}^{2}}=1$

$9{x}^{2}-18x-16{y}^{2}+32y-151=0$

$y=\frac{3}{4}\left(x-1\right)+1,y=-\frac{3}{4}\left(x-1\right)+1$

$16{y}^{2}+96y-4{x}^{2}+16x+112=0$

## Graphical

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

$\frac{{x}^{2}}{49}-\frac{{y}^{2}}{16}=1$

$\frac{{x}^{2}}{64}-\frac{{y}^{2}}{4}=1$

$\frac{{y}^{2}}{9}-\frac{{x}^{2}}{25}=1$

$81{x}^{2}-9{y}^{2}=1$

$\frac{{\left(y+5\right)}^{2}}{9}-\frac{{\left(x-4\right)}^{2}}{25}=1$

$\frac{{\left(x-2\right)}^{2}}{8}-\frac{{\left(y+3\right)}^{2}}{27}=1$

$\frac{{\left(y-3\right)}^{2}}{9}-\frac{{\left(x-3\right)}^{2}}{9}=1$

$-4{x}^{2}-8x+16{y}^{2}-32y-52=0$

${x}^{2}-8x-25{y}^{2}-100y-109=0$

$-{x}^{2}+8x+4{y}^{2}-40y+88=0$

$64{x}^{2}+128x-9{y}^{2}-72y-656=0$

$16{x}^{2}+64x-4{y}^{2}-8y-4=0$

$-100{x}^{2}+1000x+{y}^{2}-10y-2575=0$

$4{x}^{2}+16x-4{y}^{2}+16y+16=0$

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

Vertices at $\text{\hspace{0.17em}}\left(3,0\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(-3,0\right)\text{\hspace{0.17em}}$ and one focus at $\text{\hspace{0.17em}}\left(5,0\right).$

$\frac{{x}^{2}}{9}-\frac{{y}^{2}}{16}=1$

Vertices at $\text{\hspace{0.17em}}\left(0,6\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(0,-6\right)\text{\hspace{0.17em}}$ and one focus at $\text{\hspace{0.17em}}\left(0,-8\right).$

Vertices at $\text{\hspace{0.17em}}\left(1,1\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(11,1\right)\text{\hspace{0.17em}}$ and one focus at $\text{\hspace{0.17em}}\left(12,1\right).$

$\frac{{\left(x-6\right)}^{2}}{25}-\frac{{\left(y-1\right)}^{2}}{11}=1$

Center: $\text{\hspace{0.17em}}\left(0,0\right);$ vertex: $\text{\hspace{0.17em}}\left(0,-13\right);$ one focus: $\text{\hspace{0.17em}}\left(0,\sqrt{313}\right).$

Center: $\text{\hspace{0.17em}}\left(4,2\right);$ vertex: $\text{\hspace{0.17em}}\left(9,2\right);$ one focus: $\text{\hspace{0.17em}}\left(4+\sqrt{26},2\right).$

$\frac{{\left(x-4\right)}^{2}}{25}-\frac{{\left(y-2\right)}^{2}}{1}=1$

Center: $\text{\hspace{0.17em}}\left(3,5\right);\text{\hspace{0.17em}}$ vertex: $\text{\hspace{0.17em}}\left(3,11\right);\text{\hspace{0.17em}}$ one focus: $\text{\hspace{0.17em}}\left(3,5+2\sqrt{10}\right).$

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

$\frac{{y}^{2}}{16}-\frac{{x}^{2}}{25}=1$

$\frac{{y}^{2}}{9}-\frac{{\left(x+1\right)}^{2}}{9}=1$

$\frac{{\left(x+3\right)}^{2}}{25}-\frac{{\left(y+3\right)}^{2}}{25}=1$

## Extensions

For the following exercises, express the equation for the hyperbola as two functions, with $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ as a function of $\text{\hspace{0.17em}}x.\text{\hspace{0.17em}}$ Express as simply as possible. Use a graphing calculator to sketch the graph of the two functions on the same axes.

$\frac{{x}^{2}}{4}-\frac{{y}^{2}}{9}=1$

$\frac{{y}^{2}}{9}-\frac{{x}^{2}}{1}=1$

$y\left(x\right)=3\sqrt{{x}^{2}+1},y\left(x\right)=-3\sqrt{{x}^{2}+1}$

$\frac{{\left(x-2\right)}^{2}}{16}-\frac{{\left(y+3\right)}^{2}}{25}=1$

$-4{x}^{2}-16x+{y}^{2}-2y-19=0$

$y\left(x\right)=1+2\sqrt{{x}^{2}+4x+5},y\left(x\right)=1-2\sqrt{{x}^{2}+4x+5}$

$4{x}^{2}-24x-{y}^{2}-4y+16=0$

## 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 and its closest distance to the center fountain is 5 yards.

$\frac{{x}^{2}}{25}-\frac{{y}^{2}}{25}=1$

The hedge will follow the asymptotes and its closest distance to the center fountain is 6 yards.

The hedge will follow the asymptotes $\text{\hspace{0.17em}}y=\frac{1}{2}x\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}y=-\frac{1}{2}x,$ and its closest distance to the center fountain is 10 yards.

$\frac{{x}^{2}}{100}-\frac{{y}^{2}}{25}=1$

The hedge will follow the asymptotes $\text{\hspace{0.17em}}y=\frac{2}{3}x\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}y=-\frac{2}{3}x,$ and its closest distance to the center fountain is 12 yards.

The hedge will follow the asymptotes and its closest distance to the center fountain is 20 yards.

$\frac{{x}^{2}}{400}-\frac{{y}^{2}}{225}=1$

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 $\text{\hspace{0.17em}}y=x-2\text{\hspace{0.17em}}$ 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 $\text{\hspace{0.17em}}y=-x+2.\text{\hspace{0.17em}}$

The object enters along a path approximated by the line $\text{\hspace{0.17em}}y=2x-2\text{\hspace{0.17em}}$ 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 $\text{\hspace{0.17em}}y=-2x+2.\text{\hspace{0.17em}}$

$\frac{{\left(x-1\right)}^{2}}{0.25}-\frac{{y}^{2}}{0.75}=1$

The object enters along a path approximated by the line $\text{\hspace{0.17em}}y=0.5x+2\text{\hspace{0.17em}}$ 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 $\text{\hspace{0.17em}}y=-0.5x-2.\text{\hspace{0.17em}}$

The object enters along a path approximated by the line $\text{\hspace{0.17em}}y=\frac{1}{3}x-1\text{\hspace{0.17em}}$ 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

$\frac{{\left(x-3\right)}^{2}}{4}-\frac{{y}^{2}}{5}=1$

The object It enters along a path approximated by the line $\text{\hspace{0.17em}}y=3x-9\text{\hspace{0.17em}}$ 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 $\text{\hspace{0.17em}}y=-3x+9.\text{\hspace{0.17em}}$

x=-b+_Гb2-(4ac) ______________ 2a
I've run into this: x = r*cos(angle1 + angle2) Which expands to: x = r(cos(angle1)*cos(angle2) - sin(angle1)*sin(angle2)) The r value confuses me here, because distributing it makes: (r*cos(angle2))(cos(angle1) - (r*sin(angle2))(sin(angle1)) How does this make sense? Why does the r distribute once
so good
abdikarin
this is an identity when 2 adding two angles within a cosine. it's called the cosine sum formula. there is also a different formula when cosine has an angle minus another angle it's called the sum and difference formulas and they are under any list of trig identities
How can you tell what type of parent function a graph is ?
generally by how the graph looks and understanding what the base parent functions look like and perform on a graph
William
if you have a graphed line, you can have an idea by how the directions of the line turns, i.e. negative, positive, zero
William
y=x will obviously be a straight line with a zero slope
William
y=x^2 will have a parabolic line opening to positive infinity on both sides of the y axis vice versa with y=-x^2 you'll have both ends of the parabolic line pointing downward heading to negative infinity on both sides of the y axis
William
y=x will be a straight line, but it will have a slope of one. Remember, if y=1 then x=1, so for every unit you rise you move over positively one unit. To get a straight line with a slope of 0, set y=1 or any integer.
Aaron
yes, correction on my end, I meant slope of 1 instead of slope of 0
William
what is f(x)=
I don't understand
Joe
Typically a function 'f' will take 'x' as input, and produce 'y' as output. As 'f(x)=y'. According to Google, "The range of a function is the complete set of all possible resulting values of the dependent variable (y, usually), after we have substituted the domain."
Thomas
Sorry, I don't know where the "Â"s came from. They shouldn't be there. Just ignore them. :-)
Thomas
Darius
Thanks.
Thomas
Â
Thomas
It is the Â that should not be there. It doesn't seem to show if encloses in quotation marks. "Â" or 'Â' ... Â
Thomas
Now it shows, go figure?
Thomas
what is this?
i do not understand anything
unknown
lol...it gets better
Darius
I've been struggling so much through all of this. my final is in four weeks 😭
Tiffany
this book is an excellent resource! have you guys ever looked at the online tutoring? there's one that is called "That Tutor Guy" and he goes over a lot of the concepts
Darius
thank you I have heard of him. I should check him out.
Tiffany
is there any question in particular?
Joe
I have always struggled with math. I get lost really easy, if you have any advice for that, it would help tremendously.
Tiffany
Sure, are you in high school or college?
Darius
Hi, apologies for the delayed response. I'm in college.
Tiffany
how to solve polynomial using a calculator
So a horizontal compression by factor of 1/2 is the same as a horizontal stretch by a factor of 2, right?
The center is at (3,4) a focus is at (3,-1), and the lenght of the major axis is 26
The center is at (3,4) a focus is at (3,-1) and the lenght of the major axis is 26 what will be the answer?
Rima
I done know
Joe
What kind of answer is that😑?
Rima
I had just woken up when i got this message
Joe
Rima
i have a question.
Abdul
how do you find the real and complex roots of a polynomial?
Abdul
@abdul with delta maybe which is b(square)-4ac=result then the 1st root -b-radical delta over 2a and the 2nd root -b+radical delta over 2a. I am not sure if this was your question but check it up
Nare
This is the actual question: Find all roots(real and complex) of the polynomial f(x)=6x^3 + x^2 - 4x + 1
Abdul
@Nare please let me know if you can solve it.
Abdul
I have a question
juweeriya
hello guys I'm new here? will you happy with me
mustapha
The average annual population increase of a pack of wolves is 25.
how do you find the period of a sine graph
Period =2π if there is a coefficient (b), just divide the coefficient by 2π to get the new period
Am
if not then how would I find it from a graph
Imani
by looking at the graph, find the distance between two consecutive maximum points (the highest points of the wave). so if the top of one wave is at point A (1,2) and the next top of the wave is at point B (6,2), then the period is 5, the difference of the x-coordinates.
Am
you could also do it with two consecutive minimum points or x-intercepts
Am
I will try that thank u
Imani
Case of Equilateral Hyperbola
ok
Zander
ok
Shella
f(x)=4x+2, find f(3)
Benetta
f(3)=4(3)+2 f(3)=14
lamoussa
14
Vedant
pre calc teacher: "Plug in Plug in...smell's good" f(x)=14
Devante
8x=40
Chris
Explain why log a x is not defined for a < 0
the sum of any two linear polynomial is what
Momo
how can are find the domain and range of a relations
the range is twice of the natural number which is the domain
Morolake