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McGraw Hill Glencoe Algebra 2, 2012 View details

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Exercise 39 Page 648

If necessary, rewrite the equation. Then identify whether the ellipse is vertical or horizontal.

Center: (0,4)
Foci: (4sqrt(2),4 ) and (- 4sqrt(2),4)
Length of the Major Axis: 12 units
Length of the Minor Axis: 4 units
Graph:

Practice makes perfect

First, we need to rewrite this equation such that it fits the general equation of an ellipse. Horizontal Ellipse: & (x- h)^2/a^2 + (y- k)^2/b^2 = 1 [0.8em] Vertical Ellipse: & (y- k)^2/a^2 + (x- h)^2/b^2 = 1 Note that the denominator of the x-variable is greater than the denominator of the y-variable. Therefore, the equation represents a horizontal ellipse. Let's rewrite the equation a bit to make it easier to identify the values of h, k, a, and b.

x^2/36+(y-4)^2/4=1

WriteDiff

Write as a difference

(x-0)^2/36+(y-4)^2/4=1

WritePow

Write as a power

(x- 0)^2/6^2+(y- 4)^2/2^2=1

Let's now use the equation to find the desired information.

Equation of the Horizontal Ellipse	(x- h)^2/a^2+(y- k)^2/b^2=1, a and b positive, with a>b	(x- 0)^2/6^2+(y- 4)^2/2^2=1
Center	( h, k)	( 0, 4)
Length of Major Axis	2 a units	2( 6)=12 units
Length of Minor Axis	2b units	2(2)=4 units
Foci	( h± c, k), c^2= a^2-b^2	( 0± c, 4), c^2= 6^2-2^2

Finally, let's calculate the value of c and find the foci.

c^2= 6^2-2^2

▼

Solve for c

CalcPow

Calculate power

c^2=36-4

SubTerm

Subtract term

c^2=32

SqrtEqn

sqrt(LHS)=sqrt(RHS)

c=±sqrt(32)

SplitIntoFactors

Split into factors

c=±sqrt(16(2))

RootProd

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

c=± 4sqrt(2)

Now we can add and subtract c to find the foci. Foci (0± 4sqrt(2),4) ⇓ (4sqrt(2),4) and (- 4sqrt(2),4) We will now use the information we found to draw the ellipse.

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Exercises p.646-650

Exercise 39 Page 648

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