# 10.1 The ellipse  (Page 9/16)

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$4{x}^{2}+16{y}^{2}=1$

$\frac{{\left(x-2\right)}^{2}}{49}+\frac{{\left(y-4\right)}^{2}}{25}=1$

$\frac{{\left(x-2\right)}^{2}}{{7}^{2}}+\frac{{\left(y-4\right)}^{2}}{{5}^{2}}=1;\text{\hspace{0.17em}}$ Endpoints of major axis $\text{\hspace{0.17em}}\left(9,4\right),\left(-5,4\right).\text{\hspace{0.17em}}$ Endpoints of minor axis $\text{\hspace{0.17em}}\left(2,9\right),\left(2,-1\right).\text{\hspace{0.17em}}$ Foci at $\text{\hspace{0.17em}}\left(2+2\sqrt{6},4\right),\left(2-2\sqrt{6},4\right).$

$\frac{{\left(x-2\right)}^{2}}{81}+\frac{{\left(y+1\right)}^{2}}{16}=1$

$\frac{{\left(x+5\right)}^{2}}{4}+\frac{{\left(y-7\right)}^{2}}{9}=1$

$\frac{{\left(x+5\right)}^{2}}{{2}^{2}}+\frac{{\left(y-7\right)}^{2}}{{3}^{2}}=1;\text{\hspace{0.17em}}$ Endpoints of major axis $\text{\hspace{0.17em}}\left(-5,10\right),\left(-5,4\right).\text{\hspace{0.17em}}$ Endpoints of minor axis $\text{\hspace{0.17em}}\left(-3,7\right),\left(-7,7\right).\text{\hspace{0.17em}}$ Foci at $\text{\hspace{0.17em}}\left(-5,7+\sqrt{5}\right),\left(-5,7-\sqrt{5}\right).$

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

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

$\frac{{\left(x-1\right)}^{2}}{{3}^{2}}+\frac{{\left(y-4\right)}^{2}}{{2}^{2}}=1;\text{\hspace{0.17em}}$ Endpoints of major axis $\text{\hspace{0.17em}}\left(4,4\right),\left(-2,4\right).\text{\hspace{0.17em}}$ Endpoints of minor axis $\text{\hspace{0.17em}}\left(1,6\right),\left(1,2\right).\text{\hspace{0.17em}}$ Foci at $\text{\hspace{0.17em}}\left(1+\sqrt{5},4\right),\left(1-\sqrt{5},4\right).$

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

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

$\frac{{\left(x-3\right)}^{2}}{{\left(3\sqrt{2}\right)}^{2}}+\frac{{\left(y-5\right)}^{2}}{{\left(\sqrt{2}\right)}^{2}}=1;\text{\hspace{0.17em}}$ Endpoints of major axis $\text{\hspace{0.17em}}\left(3+3\sqrt{2},5\right),\left(3-3\sqrt{2},5\right).\text{\hspace{0.17em}}$ Endpoints of minor axis $\text{\hspace{0.17em}}\left(3,5+\sqrt{2}\right),\left(3,5-\sqrt{2}\right).\text{\hspace{0.17em}}$ Foci at $\text{\hspace{0.17em}}\left(7,5\right),\left(-1,5\right).$

$4{x}^{2}+24x+16{y}^{2}-128y+228=0$

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

$\frac{{\left(x+5\right)}^{2}}{{\left(5\right)}^{2}}+\frac{{\left(y-2\right)}^{2}}{{\left(2\right)}^{2}}=1;\text{\hspace{0.17em}}$ Endpoints of major axis $\text{\hspace{0.17em}}\left(0,2\right),\left(-10,2\right).\text{\hspace{0.17em}}$ Endpoints of minor axis $\text{\hspace{0.17em}}\left(-5,4\right),\left(-5,0\right).\text{\hspace{0.17em}}$ Foci at $\text{\hspace{0.17em}}\left(-5+\sqrt{21},2\right),\left(-5-\sqrt{21},2\right).$

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

$4{x}^{2}+24x+25{y}^{2}+200y+336=0$

$\frac{{\left(x+3\right)}^{2}}{{\left(5\right)}^{2}}+\frac{{\left(y+4\right)}^{2}}{{\left(2\right)}^{2}}=1;\text{\hspace{0.17em}}$ Endpoints of major axis $\text{\hspace{0.17em}}\left(2,-4\right),\left(-8,-4\right).\text{\hspace{0.17em}}$ Endpoints of minor axis $\text{\hspace{0.17em}}\left(-3,-2\right),\left(-3,-6\right).\text{\hspace{0.17em}}$ Foci at $\text{\hspace{0.17em}}\left(-3+\sqrt{21},-4\right),\left(-3-\sqrt{21},-4\right).$

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

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

$\frac{{\left(x+3\right)}^{2}}{25}+\frac{{\left(y+1\right)}^{2}}{36}=1$

Foci $\text{\hspace{0.17em}}\left(-3,-1+\sqrt{11}\right),\left(-3,-1-\sqrt{11}\right)$

$\frac{{\left(x+1\right)}^{2}}{100}+\frac{{\left(y-2\right)}^{2}}{4}=1$

${x}^{2}+{y}^{2}=1$

Focus $\text{\hspace{0.17em}}\left(0,0\right)$

${x}^{2}+4{y}^{2}+4x+8y=1$

$10{x}^{2}+{y}^{2}+200x=0$

Foci $\text{\hspace{0.17em}}\left(-10,30\right),\left(-10,-30\right)$

## Graphical

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

$\frac{{x}^{2}}{25}+\frac{{y}^{2}}{36}=1$

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

Center $\text{\hspace{0.17em}}\left(0,0\right),\text{\hspace{0.17em}}$ Vertices $\text{\hspace{0.17em}}\left(4,0\right),\left(-4,0\right),\left(0,3\right),\left(0,-3\right),\text{\hspace{0.17em}}$ Foci $\text{\hspace{0.17em}}\left(\sqrt{7},0\right),\left(-\sqrt{7},0\right)$

$4{x}^{2}+9{y}^{2}=1$

$81{x}^{2}+49{y}^{2}=1$

Center $\text{\hspace{0.17em}}\left(0,0\right),\text{\hspace{0.17em}}$ Vertices $\text{\hspace{0.17em}}\left(\frac{1}{9},0\right),\left(-\frac{1}{9},0\right),\left(0,\frac{1}{7}\right),\left(0,-\frac{1}{7}\right),\text{\hspace{0.17em}}$ Foci $\text{\hspace{0.17em}}\left(0,\frac{4\sqrt{2}}{63}\right),\left(0,-\frac{4\sqrt{2}}{63}\right)$

$\frac{{\left(x-2\right)}^{2}}{64}+\frac{{\left(y-4\right)}^{2}}{16}=1$

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

Center $\text{\hspace{0.17em}}\left(-3,3\right),\text{\hspace{0.17em}}$ Vertices $\text{\hspace{0.17em}}\left(0,3\right),\left(-6,3\right),\left(-3,0\right),\left(-3,6\right),\text{\hspace{0.17em}}$ Focus $\text{\hspace{0.17em}}\left(-3,3\right)\text{\hspace{0.17em}}$

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

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

$4{x}^{2}-8x+16{y}^{2}-32y-44=0$

Center $\text{\hspace{0.17em}}\left(1,1\right),\text{\hspace{0.17em}}$ Vertices $\text{\hspace{0.17em}}\left(5,1\right),\left(-3,1\right),\left(1,3\right),\left(1,-1\right),\text{\hspace{0.17em}}$ Foci $\text{\hspace{0.17em}}\left(1,1+4\sqrt{3}\right),\left(1,1-4\sqrt{3}\right)$

${x}^{2}-8x+25{y}^{2}-100y+91=0$

${x}^{2}+8x+4{y}^{2}-40y+112=0$

Center $\text{\hspace{0.17em}}\left(-4,5\right),\text{\hspace{0.17em}}$ Vertices $\text{\hspace{0.17em}}\left(-2,5\right),\left(-6,4\right),\left(-4,6\right),\left(-4,4\right),\text{\hspace{0.17em}}$ Foci $\text{\hspace{0.17em}}\left(-4+\sqrt{3},5\right),\left(-4-\sqrt{3},5\right)$

$64{x}^{2}+128x+9{y}^{2}-72y-368=0$

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

Center $\text{\hspace{0.17em}}\left(-2,1\right),\text{\hspace{0.17em}}$ Vertices $\text{\hspace{0.17em}}\left(0,1\right),\left(-4,1\right),\left(-2,5\right),\left(-2,-3\right),\text{\hspace{0.17em}}$ Foci $\text{\hspace{0.17em}}\left(-2,1+2\sqrt{3}\right),\left(-2,1-2\sqrt{3}\right)$

$100{x}^{2}+1000x+{y}^{2}-10y+2425=0$

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

Center $\text{\hspace{0.17em}}\left(-2,-2\right),\text{\hspace{0.17em}}$ Vertices $\text{\hspace{0.17em}}\left(0,-2\right),\left(-4,-2\right),\left(-2,0\right),\left(-2,-4\right),\text{\hspace{0.17em}}$ Focus $\text{\hspace{0.17em}}\left(-2,-2\right)$

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 $\text{\hspace{0.17em}}\left(4,0\right),$ and point on graph $\text{\hspace{0.17em}}\left(0,3\right).$

Center at the origin, symmetric with respect to the x - and y -axes, focus at $\text{\hspace{0.17em}}\left(0,-2\right),$ and point on graph $\text{\hspace{0.17em}}\left(5,0\right).$

$\frac{{x}^{2}}{25}+\frac{{y}^{2}}{29}=1$

Center at the origin, symmetric with respect to the x - and y -axes, focus at $\text{\hspace{0.17em}}\left(3,0\right),$ and major axis is twice as long as minor axis.

Center $\text{\hspace{0.17em}}\left(4,2\right)$ ; vertex $\text{\hspace{0.17em}}\left(9,2\right)$ ; one focus: $\text{\hspace{0.17em}}\left(4+2\sqrt{6},2\right)$ .

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

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

Center $\text{\hspace{0.17em}}\left(-3,4\right)$ ; vertex $\text{\hspace{0.17em}}\left(1,4\right)$ ; one focus: $\text{\hspace{0.17em}}\left(-3+2\sqrt{3},4\right)$

$\frac{{\left(x+3\right)}^{2}}{16}+\frac{{\left(y-4\right)}^{2}}{4}=1$

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

$\frac{{x}^{2}}{81}+\frac{{y}^{2}}{9}=1$

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

## Extensions

For the following exercises, find the area of the ellipse. The area of an ellipse is given by the formula $\text{\hspace{0.17em}}\text{Area}=a\cdot b\cdot \pi .$

$\frac{{\left(x-3\right)}^{2}}{9}+\frac{{\left(y-3\right)}^{2}}{16}=1$

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

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

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

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

## 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.

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 $\text{\hspace{0.17em}}h,$ the height.

$\frac{{x}^{2}}{4{h}^{2}}+\frac{{y}^{2}}{\frac{1}{4}{h}^{2}}=1$

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.

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.

$\frac{{x}^{2}}{400}+\frac{{y}^{2}}{144}=1$ . Distance = 17.32 feet

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.

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

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?

#### Questions & Answers

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
Can you please help me. Tomorrow is the deadline of my assignment then I don't know how to solve that
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
divide simplify each answer 3/2÷5/4
divide simplify each answer 25/3÷5/12
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