Question

which is the standard form of the equation of the parabola that has a vertex of (-4, -6) and a directrix of y = 3? (x + 4)^2=-36(y + 6) (y + 4)^2=36(x + 6) (y - 4)^2=28(x - 6) (x - 4)^2=-28(y - 6)

Explanation:

Step1: Determine the form of the parabola

Since the directrix is a horizontal line $y = 3$, the parabola opens either up - ward or downward and its standard form is $(x - h)^2=4p(y - k)$, where $(h,k)$ is the vertex. Here, $h=-4$ and $k = - 6$.

Step2: Calculate the value of $p$

The distance between the vertex $(h,k)=(-4,-6)$ and the directrix $y = 3$ is given by $|p|$. The formula for the distance between a point $(x_0,y_0)$ and the line $y = c$ is $|y_0 - c|$. So, $|p|=|-6 - 3|=9$. Since the vertex $(-4,-6)$ is below the directrix $y = 3$, the parabola opens downward and $p=-9$.

Step3: Substitute values into the standard - form equation

Substitute $h=-4$, $k = - 6$, and $p=-9$ into $(x - h)^2=4p(y - k)$. We get $(x+4)^2=4\times(-9)(y + 6)$, which simplifies to $(x + 4)^2=-36(y + 6)$.

Answer:

$(x + 4)^2=-36(y + 6)$