Question

determine the location of the focus and the directrix of the parabola $x^{2}+8x + 8y+40 = 0$. focus: ( select , select ) directrix: select

Explanation:

Step1: Rewrite the equation in standard form

Complete the square for the $x$ - terms.
Starting with $x^{2}+8x + 8y+40 = 0$.
We have $x^{2}+8x=(x + 4)^{2}-16$.
So the equation becomes $(x + 4)^{2}-16+8y + 40=0$.
Simplify it to $(x + 4)^{2}=-8y - 24$.
Then $(x + 4)^{2}=-8(y + 3)$.

Step2: Identify the parameters

For a parabola of the form $(x - h)^{2}=4p(y - k)$, comparing with $(x + 4)^{2}=-8(y + 3)$, we have $4p=-8$, so $p=-2$, $h=-4$, and $k=-3$.

Step3: Find the focus

The focus of the parabola $(x - h)^{2}=4p(y - k)$ is at the point $(h,k + p)$.
Substituting the values, we get $h=-4$, $k=-3$, $p=-2$.
So the focus is $(-4,-3-2)=(-4,-5)$.

Step4: Find the directrix

The directrix of the parabola $(x - h)^{2}=4p(y - k)$ is given by the equation $y=k - p$.
Substituting $k=-3$ and $p=-2$, we get $y=-3-(-2)=-1$.

Answer:

Focus: $(-4,-5)$
Directrix: $y=-1$