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Pearson Algebra 2 Common Core, 2011 View details

End-of-Course Assessment

Exercise 80 Page 971

Practice makes perfect

a We are given the following equation of an ellipse.

4 x^{2} + 9 y^{2} + 8 x - 54 y + 49 = 0

We are asked to write this equation in sstandard form. Let's first remember how the standard form of an ellipse is written.

\frac{( x - h ) ^{2}}{a ^{2}} + \frac{( y - k ) ^{2}}{b ^{2}} = 1

Now we will rewrite our equation by completing the perfect square trinomials.

4 x^{2} + 9 y^{2} + 8 x - 54 y + 49 = 0

▼

Rewrite equation

AddEqn

$LHS + 36 = RHS + 36$

4 x^{2} + 9 y^{2} + 8 x - 54 y + 85 = 36

WriteSum

Write as a sum

4 x^{2} + 9 y^{2} + 8 x - 54 y + 81 + 4 = 36

CommutativePropAdd

Commutative Property of Addition

4 x^{2} + 8 x + 4 + 9 y^{2} - 54 y + 81 = 36

FactorOut

Factor out $4$

4 (x^{2} + 2 x + 1) + 9 y^{2} - 54 y + 81 = 36

FacPosPerfectSquare

$a^{2} + 2 a b + b^{2} = (a + b)^{2}$

4 (x + 1)^{2} + 9 y^{2} - 54 y + 81 = 36

FactorOut

Factor out $9$

4 (x + 1)^{2} + 9 (y^{2} - 6 x + 9) = 36

FacNegPerfectSquare

$a^{2} - 2 a b + b^{2} = (a - b)^{2}$

4 (x + 1)^{2} + 9 (y - 3)^{2} = 36

DivEqn

$LHS / 36 = RHS / 36$

\frac{4 ( x + 1 ) ^{2} + 9 ( y - 3 ) ^{2}}{3 6} = 1

WriteSumFrac

Write as a sum of fractions

\frac{4 ( x + 1 ) ^{2}}{3 6} + \frac{9 ( y - 3 ) ^{2}}{3 6} = 1

CancelCommonFac

Cancel out common factors

\frac{( x + 1 ) ^{2}}{9} + \frac{( y - 3 ) ^{2}}{4} = 1

WritePow

Write as a power

\frac{( x + 1 ) ^{2}}{3 ^{2}} + \frac{( y - 3 ) ^{2}}{2 ^{2}} = 1

$a = - (- a)$

\frac{( x - ( - 1 ) ) ^{2}}{3 ^{2}} + \frac{( y - 3 ) ^{2}}{2 ^{2}} = 1

Great! Let's now compare our obtained equation with the general form to determine the type of ellipse that our equation represents.

\frac{( x - h ) ^{2}}{a ^{2}} + ⇓ \frac{( x - ( - 1 ) ) ^{2}}{3 ^{2}} + \frac{( y - k ) ^{2}}{b ^{2}} = 1 \frac{( y - 3 ) ^{2}}{2 ^{2}} = 1

As we can see, 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.

a > b > 0 \Rightarrow 3 > 2 > 0

Therefore, our equation matches the format of a horizontal ellipse.

b We are asked to find the foci and vertices of the given ellipse. To do so, we will start by recalling the main characteristics of a horizontal ellipse.

	Horizontal Ellipse
Standard-Form Equation	$\frac{( x - h ) ^{2}}{a ^{2}} + \frac{( y - k ) ^{2}}{b ^{2}} = 1$
Center	$(h, k)$
Vertices	$(h \pm a, k)$
Co-vertices	$(h, k \pm b)$
Foci	$(h \pm c, k)$
$a, b, c$ relationship, $a > b > 0$	$c^{2} = a^{2} - b^{2}$

Now, let's consider our equation in standard form one more time.

\frac{( x - ( - 1 ) ) ^{2}}{3 ^{2}} + \frac{( y - 3 ) ^{2}}{2 ^{2}} = 1

We see that

a = 3,

b = 2,

h = - 1,

and

k = 3 .

The only value we do not know is

c .

Let's find it!

c^{2} = a^{2} - b^{2}

SubstituteII

$a = 3$ , $b = 2$

c^{2} = 3^{2} - 2^{2}

▼

Solve for

c

CalcPow

Calculate power

c^{2} = 9 - 4

SubTerm

Subtract term

c^{2} = 5

SqrtEqn

$LHS = RHS$

c = 5

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.

Foci	Vertices
$(- 1 \pm 6, 3)$ $⇓$ $(- 1 + 6, 3) and (- 1 - 6, 3)$	$(- 1 \pm 3, 3)$ $⇓$ $(2, 3) and (- 4, 3)$

c Last, we will graph this horizontal ellipse. Since we already know the foci and vertices of the ellipse, let's now find the center and co-vertices.

Center	Co-vertices
$(- 1, 3)$	$(- 1, 3 \pm 2,)$ $⇓$ $(- 1, 5)$ and $(- 1, 1)$

To graph the ellipse we plot the center, vertices, and co-vertices. Then we connect the vertices and co-vertices with a smooth curve. Let's do it!

Graph of ellipse (x+1)^2/9+(y-3)^2/4=1. The two vertices, two covertices, and the center is labeled.

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Exercises p.964-971

Exercise 80 Page 971

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