Let's rewrite the given equation in order to identify the .
4x^2+y^2=16
4x^2+y^2/16=1
4x^2/16+y^2/16=1
x^2/4+y^2/16=1
x^2/2^2+y^2/4^2=1
(x-0)^2/2^2+y^2/4^2=1
(x-0)^2/2^2+(y-0)^2/4^2=1
We can see that the containing the are both raised to the power of 2 and are both positive. Moreover, the denominator of the binomial containing the y-variable is greater than the denominator of the binomial that contains the x-variable. Therefore, our equation matches the format of a vertical .
Graphing the Ellipse
Let's recall the main characteristics of this type of ellipse.
|
|
Vertical Ellipse
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| Standard-Form Equation
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(x- h)^2/b^2+(y-k)^2/a^2=1
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| Center
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( h,k)
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| Vertices
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( h,k± a)
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| Co-vertices
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( h± b,k)
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| Foci
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( h,k± c)
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| a,b,c relationship, a>b>0
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c^2= a^2- b^2
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Consider our equation one more time.
(x- 0)^2/2^2+(y- )^2/4^2=1
We see that a= 4, b= 2, h= 0, and k= . The only value we do not know is c. Let's find it!
c^2=a^2-b^2
c^2= 4^2- 2^2
c^2=16-4
c^2=12
c=sqrt(12)
c=sqrt(4(3))
c=2sqrt(3)
Note that we only took the , because to find the foci we will add and subtract the value of c. Therefore, its sign is irrelevant. We can now write the desired information.
| Center
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Foci
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| ( 0, )
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( 0 , ± 2sqrt(3))
⇓
( 0,2sqrt(3) ) and ( 0,- 2sqrt(3) )
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We already know the center and the foci of the ellipse. Let's find the vertices and the co-vertices.
| Vertices
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Co-vertices
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( 0, ± 4)
⇓
(0,4) and (0,- 4)
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( 0± 2, )
⇓
(2,0) and (- 2,0)
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To graph the ellipse, we plot the vertices and co-vertices. Then, we connect them with a smooth curve.
