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Complete the square in the given equation to obtain the standard form and identify the conic section.
Equation in Standard Form: (x-0)^2/4^2+(y-0)^2/3^2= 1
Conic Section: Ellipse
Graph:
Let's rewrite the given equation in order to identify the conic section.
We can see that the binomials containing the variables are both raised to the power of 2 and are both positive. Moreover, the denominator of the binomial containing the x-variable is greater than the denominator of the binomial that contains the y-variable. Therefore, our equation matches the format of a horizontal ellipse.
Let's recall the main characteristics of this type of ellipse.
| Horizontal Ellipse | |
|---|---|
| Standard-Form Equation | (x- h)^2/a^2+(y-k)^2/b^2=1 |
| Center | ( h,k) |
| Vertices | ( h± a,k) |
| Co-vertices | ( h,k± b) |
| Foci | ( h± c,k) |
| a,b,c relationship, a>b>0 | c^2= a^2- b^2 |
Consider our equation one more time. (x- 0)^2/4^2+(y- )^2/3^2=1 We see that a= 4, b= 3, h= 0, and k= . The only value we do not know is c. Let's find it!
Let's identify the center and the foci.
| Center | Foci |
|---|---|
| ( 0, ) | ( 0 , ± sqrt(7)) ⇓ ( 0,sqrt(7) ) and ( 0,- sqrt(7) ) |
We already know the center and the foci of the ellipse. Let's find the vertices and the co-vertices.
| Vertices | Co-vertices |
|---|---|
| ( 0± 4, ) ⇓ (4,0) and (- 4,0) |
( 0, ± 3) ⇓ (0,3) and (0,- 3) |
To graph the ellipse, we plot the vertices and co-vertices. Then, we connect them with a smooth curve.