Exercise 44 Page 648

To find the desired information about the ellipse, we need to rewrite the given equation by completing the square. We will need to complete the square twice, once for each variable. Let's start by rearranging the equation and factoring out any common factors. Remember, to complete a square we need to add and subtract the value of ${\left(\frac{b}{2}\right)}^2.$ In this case, we need to do it for the linear coefficients of the $x\text{-}$ and $y\text{-}$terms. Let's do it!

$7\left(x^2-4x\right)+3\left(y^2-4y\right)=\text{-}19$

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Complete the square

CompleteSquare

Complete the square

$7(\left(x^2-4x+2^2\right)-2^2)+3(\left(y^2-4y+2^2\right)-2^2)=\text{-}19$

SplitIntoFactors

Split into factors

$7(\left(x^2-2x(2)+2^2\right)-2^2)+3(\left(y^2-2y(2)+2^2\right)-2^2)=\text{-}19$

FacNegPerfectSquare

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

$7((x-2{)}^2-2^2)+3((y-2{)}^2-2^2)=\text{-}19$

CalcPow

Calculate power

$7((x-2{)}^2-4)+3((y-2{)}^2-4)=\text{-}19$

Next, we need to rewrite this equation such that it fits the general equation of an ellipse.

$$\begin{array}{rl}\text{Horizontal Ellipse:}& \frac{(x-h{)}^2}{a^2}+\frac{(y-k{)}^2}{b^2}=1\\ \text{Vertical Ellipse:}& \frac{(y-k{)}^2}{a^2}+\frac{(x-h{)}^2}{b^2}=1\end{array}$$

We will need to use the Distributive Property first and then we can continue with the Properties of Equality.

$7((x-2{)}^2-4)+3((y-2{)}^2-4)=\text{-}19$

Distr

Distribute $7\&3$

$7(x-2{)}^2-28+3(y-2{)}^2-12=\text{-}19$

AddEqn

$\text{LHS}+28=\text{RHS}+28$

$7(x-2{)}^2+3(y-2{)}^2-12=9$

AddEqn

$\text{LHS}+12=\text{RHS}+12$

$7(x-2{)}^2+3(y-2{)}^2=21$

DivEqn

$\text{LHS}/21=\text{RHS}/21$

$\frac{7(x-2{)}^2+3(y-2{)}^2}{21}=1$

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Simplify left-hand side

WriteSumFrac

Write as a sum of fractions

$\frac{7(x-2{)}^2}{21}+\frac{3(y-2{)}^2}{21}=1$

SimpQuot

Simplify quotient

$\frac{(x-2{)}^2}{3}+\frac{(y-2{)}^2}{7}=1$

Because the denominator of the $y\text{-}$variable is greater than the denominator of the $x\text{-}$variable, the equation represents a vertical ellipse. Let's rewrite the equation a bit to make it easier to identify the values of $h,$ $k,$ $a,$ and $b.$

$\frac{(x-2{)}^2}{3}+\frac{(y-2{)}^2}{7}=1$

$a={\left(\sqrt{a}\right)}^2$

$\frac{(x-2{)}^2}{{\left(\sqrt{3}\right)}^2}+\frac{(y-2{)}^2}{{\left(\sqrt{7}\right)}^2}=1$

CommutativePropAdd

Commutative Property of Addition

$\frac{(y-2{)}^2}{{\left(\sqrt{7}\right)}^2}+\frac{(x-2{)}^2}{{\left(\sqrt{3}\right)}^2}=1$

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

Equation of the Vertical Ellipse	$\frac{(y-k{)}^2}{a^2}+\frac{(x-h{)}^2}{b^2}=1,$ $a$ and $b$ positive, with $a>b$	$\frac{(y-2{)}^2}{{\left(\sqrt{7}\right)}^2}+\frac{(x-2{)}^2}{{\left(\sqrt{3}\right)}^2}=1$
Center	$(h,k)$	$(2,2)$
Length of Major Axis	$2a$ units	$2\sqrt{7}$ units
Length of Minor Axis	$2b$ units	$2\sqrt{3}$ units
Foci	$(h,k\pm c),$ $c^2=a^2-b^2$	$(2,2\pm c),$ $c^2={\left(\sqrt{7}\right)}^2-{\left(\sqrt{3}\right)}^2$

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

$c^2={\left(\sqrt{7}\right)}^2-{\left(\sqrt{3}\right)}^2$

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Solve for $c$

PowSqrt

${\left(\sqrt{a}\right)}^2=a$

$c^2=7-3$

SubTerm

Subtract term

$c^2=4$

RadicalEqn

$\sqrt[]{\text{LHS}}=\sqrt[]{\text{RHS}}$

$c=\pm 2$

Now we can add and subtract $c$ to find the foci.

$$\begin{array}{c}\underline{\text{Foci}}\\ \left(2,2\pm 2\right)\\ \Downarrow \\ (2,4)\text{ and }(2,0)\end{array}$$

We will now use the information we found to draw the ellipse.