Question

complete the square to find and write the standard form for the ellipse $x^{2}+8x + 4y^{2}-16y = 0$

Explanation:

Step1: Complete the square for x - terms

For the $x$ - terms in the equation $x^{2}+8x + 4y^{2}-16y=0$, we use the formula $(a + b)^2=a^{2}+2ab + b^{2}$. In $x^{2}+8x$, if $a = x$ and $2ab=8x$, then $b = 4$. So $x^{2}+8x=(x + 4)^{2}-16$.

Step2: Complete the square for y - terms

For the $y$ - terms $4y^{2}-16y$, first factor out the coefficient of $y^{2}$: $4(y^{2}-4y)$. For $y^{2}-4y$, using the formula $(a + b)^2=a^{2}+2ab + b^{2}$ with $a = y$ and $2ab=-4y$, then $b=-2$. So $y^{2}-4y=(y - 2)^{2}-4$, and $4(y^{2}-4y)=4((y - 2)^{2}-4)=4(y - 2)^{2}-16$.

Step3: Rewrite the original equation

Substitute the completed - square expressions back into the original equation:

$$
LATEXBLOCK0
$$

Step4: Write in standard form

Divide the entire equation by 32 to get the standard form of the ellipse $\frac{(x + 4)^{2}}{32}+\frac{4(y - 2)^{2}}{32}=1$, which simplifies to $\frac{(x + 4)^{2}}{32}+\frac{(y - 2)^{2}}{8}=1$.

Answer:

$\frac{(x + 4)^{2}}{32}+\frac{(y - 2)^{2}}{8}=1$