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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

Cos45/sec30+cosec30=
dinesh Reply
Cos 45 = 1/ √ 2 sec 30 = 2/√3 cosec 30 = 2. =1/√2 / 2/√3+2 =1/√2/2+2√3/√3 =1/√2*√3/2+2√3 =√3/√2(2+2√3) =√3/2√2+2√6 --------- (1) =√3 (2√6-2√2)/((2√6)+2√2))(2√6-2√2) =2√3(√6-√2)/(2√6)²-(2√2)² =2√3(√6-√2)/24-8 =2√3(√6-√2)/16 =√18-√16/8 =3√2-√6/8 ----------(2)
exercise 1.2 solution b....isnt it lacking
Miiro Reply
I dnt get dis work well
john Reply
what is one-to-one function
Iwori Reply
what is the procedure in solving quadratic equetion at least 6?
Qhadz Reply
Almighty formula or by factorization...or by graphical analysis
Damian
I need to learn this trigonometry from A level.. can anyone help here?
wisdom Reply
yes am hia
Miiro
tanh2x =2tanhx/1+tanh^2x
Gautam Reply
cos(a+b)+cos(a-b)/sin(a+b)-sin(a-b)=cotb ... pls some one should help me with this..thanks in anticipation
favour Reply
f(x)=x/x+2 given g(x)=1+2x/1-x show that gf(x)=1+2x/3
Ken Reply
proof
AUSTINE
sebd me some questions about anything ill solve for yall
Manifoldee Reply
cos(a+b)+cos(a-b)/sin(a+b)-sin(a-b)= cotb
favour
how to solve x²=2x+8 factorization?
Kristof Reply
x=2x+8 x-2x=2x+8-2x x-2x=8 -x=8 -x/-1=8/-1 x=-8 prove: if x=-8 -8=2(-8)+8 -8=-16+8 -8=-8 (PROVEN)
Manifoldee
x=2x+8
Manifoldee
×=2x-8 minus both sides by 2x
Manifoldee
so, x-2x=2x+8-2x
Manifoldee
then cancel out 2x and -2x, cuz 2x-2x is obviously zero
Manifoldee
so it would be like this: x-2x=8
Manifoldee
then we all know that beside the variable is a number (1): (1)x-2x=8
Manifoldee
so we will going to minus that 1-2=-1
Manifoldee
so it would be -x=8
Manifoldee
so next step is to cancel out negative number beside x so we get positive x
Manifoldee
so by doing it you need to divide both side by -1 so it would be like this: (-1x/-1)=(8/-1)
Manifoldee
so -1/-1=1
Manifoldee
so x=-8
Manifoldee
SO THE ANSWER IS X=-8
Manifoldee
so we should prove it
Manifoldee
x=2x+8 x-2x=8 -x=8 x=-8 by mantu from India
mantu
lol i just saw its x²
Manifoldee
x²=2x-8 x²-2x=8 -x²=8 x²=-8 square root(x²)=square root(-8) x=sq. root(-8)
Manifoldee
I mean x²=2x+8 by factorization method
Kristof
I think x=-2 or x=4
Kristof
x= 2x+8 ×=8-2x - 2x + x = 8 - x = 8 both sides divided - 1 -×/-1 = 8/-1 × = - 8 //// from somalia
Mohamed
i am in
Cliff
1KI POWER 1/3 PLEASE SOLUTIONS
Prashant Reply
hii
Amit
how are you
Dorbor
well
Biswajit
can u tell me concepts
Gaurav
Find the possible value of 8.5 using moivre's theorem
Reuben Reply
which of these functions is not uniformly cintinuous on (0, 1)? sinx
Pooja Reply
helo
Akash
hlo
Akash
Hello
Hudheifa
which of these functions is not uniformly continuous on 0,1
Basant Reply
Practice Key Terms 7

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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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