Exercise 1 Page 634

Let's rewrite the given equation in order to identify the conic section. We will do it by completing the square. 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.

Graphing the 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- 3)^2/6^2+(y-( - 2))^2/3^2=1 We see that a= 6, b= 3, h= 3, and k= - 2. The only value we do not know is c. Let's find it!

c^2=a^2-b^2

SubstituteII

b= 3, a= 6

c^2= 6^2- 3^2

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Solve for c

CalcPow

Calculate power

c^2=36-9

SubTerm

Subtract term

c^2=27

RadicalEqn

sqrt(LHS)=sqrt(RHS)

c=sqrt(27)

SplitIntoFactors

Split into factors

c=sqrt(9(3))

SqrtProd

sqrt(a* b)=sqrt(a)*sqrt(b)

c=3sqrt(3)

Note that we only took the principal root, 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	Foci
( 3, - 2)	( 3 ± 3sqrt(3), - 2) ⇓ ( 3+3sqrt(3), - 2 ) and ( 3-3sqrt(3),- 2 )

Let's find the vertices and the co-vertices.

Vertices	Co-vertices
( 3± 6, - 2) ⇓ (9,- 2) and (- 3,- 2)	( 3, - 2± 3) ⇓ (3,1) and (3,- 5)

To graph the ellipse, we plot the vertices and co-vertices. Then, we connect them with a smooth curve.