McGraw Hill Glencoe Algebra 2, 2012
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McGraw Hill Glencoe Algebra 2, 2012 View details
Study Guide and Review
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Exercise 53 Page 649

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:

Practice makes perfect

Let's rewrite the given equation in order to identify the conic section.

9x^2+16y^2=144
â–¼
Rewrite equation
9x^2/144+16y^2/144=1
x^2/16+y^2/9=1
x^2/4^2+y^2/3^2=1
(x- 0)^2/4^2+(y- )^2/3^2=1

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.

Finding the Key Features

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!

c^2=a^2-b^2
c^2= 4^2- 3^2
â–¼
Solve for c
c^2=16-9
c^2=7
c=± sqrt(7)

Let's identify the center and the foci.

Center Foci
( 0, ) ( 0 , ± sqrt(7))
⇓
( 0,sqrt(7) ) and ( 0,- sqrt(7) )

Graphing the Ellipse

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.