Question

The equation of directrix of parabola \(x^2 + 8x + 12y + 4 = 0\) is

• A. \(y + 4 = 0\)
• B. \(y - 1 = 0\)
• C. \(y - 4 = 0\)
• D. \(y - 2 = 0\)

Hint:

To find directrix, convert parabola to standard form, identify vertex and p, then use formula directrix: y = k - p (for vertical parabola).

Answer 1

The Correct Option is C

Step 1: Complete the square in x.
\[x^2+8x+12y+4=0 \implies (x+4)^2 - 16 + 12y+4=0 \implies (x+4)^2 = -12y+12 = -12(y-1)\]

Step 2: Match the standard form.
Comparing with \((x-h)^2=4p(y-k)\), we get vertex \((h,k)=(-4,1)\) and \(4p=-12 \implies p=-3\). The negative p means the parabola opens downward.

Step 3: Locate the directrix.
For a parabola opening downward from vertex \((h,k)\) with parameter p, the directrix sits at \(y = k - p\), so
\[y = 1-(-3) = 4\]

Step 4: Conclusion.
\[\boxed{y-4=0}\]