Question:
The equation of directrix of parabola \(x^2 + 8x + 12y + 4 = 0\) is
The equation of directrix of parabola \(x^2 + 8x + 12y + 4 = 0\) is
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To find directrix, convert parabola to standard form, identify vertex and p, then use formula directrix: y = k - p (for vertical parabola).
Updated On: Jul 18, 2026
- \(y + 4 = 0\)
- \(y - 1 = 0\)
- \(y - 4 = 0\)
- \(y - 2 = 0\)
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The Correct Option is C
Solution and Explanation
Step 1: Standard form of parabola.
Given: \(x^2 + 8x + 12y + 4 = 0\)
Complete the square: \(x^2 + 8x = (x+4)^2 - 16\)
Step 2: Rewrite equation.
\((x+4)^2 -16 + 12y +4 = 0 \implies (x+4)^2 + 12y -12 = 0 \implies (x+4)^2 = -12(y - 1)\) ??? Check carefully: \( (x+4)^2 = -12y + 12 \implies (x+4)^2 = -12(y -1) \) correct.
Step 3: Parabola parameters.
Standard form \( (x-h)^2 = 4p(y-k) \), compare: \(4p = -12 \implies p=-3\), vertex \((-4,1)\)
Step 4: Directrix.
Directrix \(y = k - p = 1 - (-3) = 4\)
Step 5: Final conclusion.
Hence, directrix equation is \[ \boxed{y - 4 = 0} \]
Given: \(x^2 + 8x + 12y + 4 = 0\)
Complete the square: \(x^2 + 8x = (x+4)^2 - 16\)
Step 2: Rewrite equation.
\((x+4)^2 -16 + 12y +4 = 0 \implies (x+4)^2 + 12y -12 = 0 \implies (x+4)^2 = -12(y - 1)\) ??? Check carefully: \( (x+4)^2 = -12y + 12 \implies (x+4)^2 = -12(y -1) \) correct.
Step 3: Parabola parameters.
Standard form \( (x-h)^2 = 4p(y-k) \), compare: \(4p = -12 \implies p=-3\), vertex \((-4,1)\)
Step 4: Directrix.
Directrix \(y = k - p = 1 - (-3) = 4\)
Step 5: Final conclusion.
Hence, directrix equation is \[ \boxed{y - 4 = 0} \]
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