Question

Complete the square and write in standard form. Show all work.What would be the conic section:CircleEllipseHyperbolaParabola

Answer 1

This is an ellipse. The equation is:

`((x-1)^2)/(3^2)+((y+4)^2)/(4^2)=1`

Step-by-step explanation

We have to complete the square for each variable. To do so, we have to take the first two terms and compare them with the perfect binomial squared formula,

`(a+b)^2=a^2+2ab+b^2`

For x we have to take 16x² and -32x. Since the coefficient of x is not 1, first, we have to factor out the coefficient 16,

`16x^2-32x=16(x^2-2x)`

Now, the first term of the expanded binomial would be x and the second term -2x. Thus, the binomial is,

`(x-1)^2=x^2-2x+1`

To maintain the equation, we have to subtract 1,

`16(x^2-2x+1-1)=16((x-1)^2-1)=16(x-1)^2-16`

Now, we replace (16x² - 32x) from the given equation by this equivalent expression,

`16(x-1)^2-16+9y^2+72y+16=0`

The next step is to do the same for y. We have the terms 9y² + 72y. Again, since the coefficient of y² is not 1, we factor out the coefficient 9,

`9y^2+72y=9(y^2+8y)`

Following the same reasoning as before, we have that the perfect binomial squared is,

`(y+4)^2=y^2+8y+16`

Remember to subtract the independent term to maintain the equation,

`9(y^2+8y)=9(y^2+8y+16-16)=9((y+4)^2-16)=9(y+4)^2-144`

And now, as we did for x, replace the two terms (9y² + 72y) with this equivalent expression in the equation,

`16(x-1)^2-16+9(y+4)^2-144+16=0`

Add like terms,

`\begin{gathered} 16(x-1)^2+9(y+4)^2+(-16-144+16)=0 \\ 16(x-1)^2+9(y+4)^2-144=0 \end{gathered}`

Add 144 to both sides,

`\begin{gathered} 16(x-1)^2+9(y+4)^2-144+144=0+144 \\ 16(x-1)^2+9(y+4)^2=144 \end{gathered}`

As we can see, this is the equation of an ellipse. Its standard form is,

`((x-h)^2)/(a^2)+((y-k)^2)/(b^2)=1`

So the next step is to divide both sides by 144 and also write the coefficients as fractions in the denominator,

`\begin{gathered} (16(x-1)^2)/(144)+(9(y+4)^2)/(144)=(144)/(144) \\ \\ ((x-1)^2)/((144)/(16))+((y+4)^2)/((144)/(9))=1 \end{gathered}`

Finally, we have to write the denominators as perfect squares, so we identify the values of a and b. 144 is 12², 16 is 4² and 9 is 3²,

`((x-1)^2)/(((12)/(4))^2)+((y+4)^2)/(((12)/(3))^2)=1`

Note that we can simplify a and b,

`(12)/(4)=3\text{ and }(12)/(3)=4`

Hence, the equation of the ellipse is,

`((x-1)^2)/(3^2)+((y+4)^2)/(4^2)=1`