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Deriving the equation of an ellipse centered at the origin

To derive the equation of an ellipse    centered at the origin, we begin with the foci ( c , 0 ) and ( c , 0 ) . The ellipse is the set of all points ( x , y ) such that the sum of the distances from ( x , y ) to the foci is constant, as shown in [link] .

If ( a , 0 ) is a vertex    of the ellipse, the distance from ( c , 0 ) to ( a , 0 ) is a ( c ) = a + c . The distance from ( c , 0 ) to ( a , 0 ) is a c . The sum of the distances from the foci    to the vertex is

( a + c ) + ( a c ) = 2 a

If ( x , y ) is a point on the ellipse, then we can define the following variables:

d 1 = the distance from  ( c , 0 )  to  ( x , y ) d 2 = the distance from  ( c , 0 )  to  ( x , y )

By the definition of an ellipse, d 1 + d 2 is constant for any point ( x , y ) on the ellipse. We know that the sum of these distances is 2 a for the vertex ( a , 0 ) . It follows that d 1 + d 2 = 2 a for any point on the ellipse. We will begin the derivation by applying the distance formula. The rest of the derivation is algebraic.

                                       d 1 + d 2 = ( x ( c ) ) 2 + ( y 0 ) 2 + ( x c ) 2 + ( y 0 ) 2 = 2 a Distance formula ( x + c ) 2 + y 2 + ( x c ) 2 + y 2 = 2 a Simplify expressions .                              ( x + c ) 2 + y 2 = 2 a ( x c ) 2 + y 2 Move radical to opposite side .                                ( x + c ) 2 + y 2 = [ 2 a ( x c ) 2 + y 2 ] 2 Square both sides .                      x 2 + 2 c x + c 2 + y 2 = 4 a 2 4 a ( x c ) 2 + y 2 + ( x c ) 2 + y 2 Expand the squares .                      x 2 + 2 c x + c 2 + y 2 = 4 a 2 4 a ( x c ) 2 + y 2 + x 2 2 c x + c 2 + y 2 Expand remaining squares .                                               2 c x = 4 a 2 4 a ( x c ) 2 + y 2 2 c x Combine like terms .                                     4 c x 4 a 2 = 4 a ( x c ) 2 + y 2 Isolate the radical .                                         c x a 2 = a ( x c ) 2 + y 2 Divide by 4 .                                     [ c x a 2 ] 2 = a 2 [ ( x c ) 2 + y 2 ] 2 Square both sides .                      c 2 x 2 2 a 2 c x + a 4 = a 2 ( x 2 2 c x + c 2 + y 2 ) Expand the squares .                      c 2 x 2 2 a 2 c x + a 4 = a 2 x 2 2 a 2 c x + a 2 c 2 + a 2 y 2 Distribute  a 2 .                   a 2 x 2 c 2 x 2 + a 2 y 2 = a 4 a 2 c 2 Rewrite .                     x 2 ( a 2 c 2 ) + a 2 y 2 = a 2 ( a 2 c 2 ) Factor common terms .                                x 2 b 2 + a 2 y 2 = a 2 b 2 Set  b 2 = a 2 c 2 .                              x 2 b 2 a 2 b 2 + a 2 y 2 a 2 b 2 = a 2 b 2 a 2 b 2 Divide both sides by  a 2 b 2 .                                       x 2 a 2 + y 2 b 2 = 1 Simplify .

Thus, the standard equation of an ellipse is x 2 a 2 + y 2 b 2 = 1. This equation defines an ellipse centered at the origin. If a > b , the ellipse is stretched further in the horizontal direction, and if b > a , the ellipse is stretched further in the vertical direction.

Writing equations of ellipses centered at the origin in standard form

Standard forms of equations tell us about key features of graphs. Take a moment to recall some of the standard forms of equations we’ve worked with in the past: linear, quadratic, cubic, exponential, logarithmic, and so on. By learning to interpret standard forms of equations, we are bridging the relationship between algebraic and geometric representations of mathematical phenomena.

The key features of the ellipse    are its center, vertices , co-vertices , foci    , and lengths and positions of the major and minor axes . Just as with other equations, we can identify all of these features just by looking at the standard form of the equation. There are four variations of the standard form of the ellipse. These variations are categorized first by the location of the center (the origin or not the origin), and then by the position (horizontal or vertical). Each is presented along with a description of how the parts of the equation relate to the graph. Interpreting these parts allows us to form a mental picture of the ellipse.

Questions & Answers

find the equation of the line if m=3, and b=-2
Ashley Reply
graph the following linear equation using intercepts method. 2x+y=4
ok, one moment
how do I post your graph for you?
it won't let me send an image?
also for the first one... y=mx+b so.... y=3x-2
y=mx+b you were already given the 'm' and 'b'. so.. y=3x-2
"7"has an open circle and "10"has a filled in circle who can I have a set builder notation
Fiston Reply
x=-b+_Гb2-(4ac) ______________ 2a
Ahlicia Reply
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
Carlos Reply
so good
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
consider r(a+b) = ra + rb. The a and b are the trig identity.
How can you tell what type of parent function a graph is ?
Mary Reply
generally by how the graph looks and understanding what the base parent functions look like and perform on a graph
if you have a graphed line, you can have an idea by how the directions of the line turns, i.e. negative, positive, zero
y=x will obviously be a straight line with a zero slope
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
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.
yes, correction on my end, I meant slope of 1 instead of slope of 0
what is f(x)=
Karim Reply
I don't understand
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."
Sorry, I don't know where the "Â"s came from. They shouldn't be there. Just ignore them. :-)
It is the  that should not be there. It doesn't seem to show if encloses in quotation marks. "Â" or 'Â' ... Â
Now it shows, go figure?
what is this?
unknown Reply
i do not understand anything
lol...it gets better
I've been struggling so much through all of this. my final is in four weeks 😭
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
thank you I have heard of him. I should check him out.
is there any question in particular?
I have always struggled with math. I get lost really easy, if you have any advice for that, it would help tremendously.
Sure, are you in high school or college?
Hi, apologies for the delayed response. I'm in college.
how to solve polynomial using a calculator
Ef Reply
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
Rima Reply
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?
I done know
What kind of answer is that😑?
I had just woken up when i got this message
Can you please help me. Tomorrow is the deadline of my assignment then I don't know how to solve that
i have a question.
how do you find the real and complex roots of a polynomial?
@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
This is the actual question: Find all roots(real and complex) of the polynomial f(x)=6x^3 + x^2 - 4x + 1
@Nare please let me know if you can solve it.
I have a question
hello guys I'm new here? will you happy with me
The average annual population increase of a pack of wolves is 25.
Brittany Reply
how do you find the period of a sine graph
Imani Reply
Period =2π if there is a coefficient (b), just divide the coefficient by 2π to get the new period
if not then how would I find it from a graph
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.
you could also do it with two consecutive minimum points or x-intercepts
I will try that thank u
Case of Equilateral Hyperbola
Jhon Reply
f(x)=4x+2, find f(3)
f(3)=4(3)+2 f(3)=14
pre calc teacher: "Plug in Plug in...smell's good" f(x)=14
Explain why log a x is not defined for a < 0
Baptiste Reply
the sum of any two linear polynomial is what
Esther Reply
Practice Key Terms 7

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