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? (y - 4)² = 28(x - 6); (x + 4)² = -36(y + 6); (x - 4)² = -28(y - 6); (y + 4)² = 36(x + 6)

Explanation:

Step1: Recall parabola standard form

For a parabola with vertex \((h,k)\) and directrix \(y = k - p\) (opening up/down), the standard form is \((x - h)^2 = 4p(y - k)\). If directrix is above vertex, parabola opens down (\(p<0\)).

Vertex is \((-4, -6)\), so \(h=-4\), \(k = -6\). Directrix \(y = 3\), which is above vertex (\(3 > -6\)), so parabola opens down (\(p<0\)).

Step2: Calculate \(p\)

Directrix formula: \(y = k - p\). Substitute \(y = 3\), \(k = -6\):
\(3 = -6 - p\)
Solve for \(p\): \(p = -6 - 3 = -9\).

Step3: Substitute \(h,k,p\) into formula

Standard form: \((x - h)^2 = 4p(y - k)\). Substitute \(h=-4\), \(k=-6\), \(p=-9\):
\((x - (-4))^2 = 4(-9)(y - (-6))\)
Simplify: \((x + 4)^2 = -36(y + 6)\).

Answer:

\((x + 4)^2 = -36(y + 6)\) (the middle option)