# 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 $\text{\hspace{0.17em}}y=-3x+9.\text{\hspace{0.17em}}$

can you not take the square root of a negative number
Suppose P= {-3,1,3} Q={-3,-2-1} and R= {-2,2,3}.what is the intersection
can I get some pretty basic questions
In what way does set notation relate to function notation
Ama
is precalculus needed to take caculus
It depends on what you already know. Just test yourself with some precalculus questions. If you find them easy, you're good to go.
Spiro
the solution doesn't seem right for this problem
what is the domain of f(x)=x-4/x^2-2x-15 then
x is different from -5&3
Seid
All real x except 5 and - 3
Spiro
how to prroved cos⁴x-sin⁴x= cos²x-sin²x are equal
Don't think that you can.
Elliott
how do you provided cos⁴x-sin⁴x = cos²x-sin²x are equal
What are the question marks for?
Elliott
Someone should please solve it for me Add 2over ×+3 +y-4 over 5 simplify (×+a)with square root of two -×root 2 all over a multiply 1over ×-y{(×-y)(×+y)} over ×y
For the first question, I got (3y-2)/15 Second one, I got Root 2 Third one, I got 1/(y to the fourth power) I dont if it's right cause I can barely understand the question.
Is under distribute property, inverse function, algebra and addition and multiplication function; so is a combined question
Abena
find the equation of the line if m=3, and b=-2
graph the following linear equation using intercepts method. 2x+y=4
Ashley
how
Wargod
what?
John
ok, one moment
UriEl
how do I post your graph for you?
UriEl
it won't let me send an image?
UriEl
also for the first one... y=mx+b so.... y=3x-2
UriEl
y=mx+b you were already given the 'm' and 'b'. so.. y=3x-2
Tommy
Please were did you get y=mx+b from
Abena
y=mx+b is the formula of a straight line. where m = the slope & b = where the line crosses the y-axis. In this case, being that the "m" and "b", are given, all you have to do is plug them into the formula to complete the equation.
Tommy
thanks Tommy
Nimo
0=3x-2 2=3x x=3/2 then . y=3/2X-2 I think
Given
co ordinates for x x=0,(-2,0) x=1,(1,1) x=2,(2,4)
neil
"7"has an open circle and "10"has a filled in circle who can I have a set builder notation
Where do the rays point?
Spiro
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
strategies to form the general term
carlmark
consider r(a+b) = ra + rb. The a and b are the trig identity.
Mike
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