Question

\(y=4\) is the directrix of the parabola \[ x^{2}+8x+12y+k=0. \] If \(l\) is the length of its latus rectum, then \(l-k=\) ?

• A. 4
• B. \(\sqrt8\)
• C. 12
• D. 6

Hint:

Whenever a parabola contains both \(x^2\) and \(x\), first complete the square and compare with the standard form.

Answer 1

The Correct Option is C

Concept: Standard form: \[ (x-h)^2=4a(y-k) \] Directrix: \[ y=k-a. \] Length of latus rectum: \[ l=4a. \]

Step 1: Convert parabola into standard form.
\[ x^2+8x+12y+k=0 \] \[ (x+4)^2-16+12y+k=0 \] \[ (x+4)^2=-12y+16-k. \] \[ (x+4)^2=-12\left(y-\frac{16-k}{12}\right). \] Comparing with \[ (x-h)^2=4a(y-K) \] gives \[ 4a=-12 \] \[ a=-3. \]

Step 2: Use directrix condition.
Directrix: \[ y=K-a. \] Given directrix \[ y=4. \] Hence \[ \frac{16-k}{12}-(-3)=4. \] \[ \frac{16-k}{12}+3=4. \] \[ \frac{16-k}{12}=1. \] \[ 16-k=12. \] \[ k=4. \]

Step 3: Find latus rectum length.
\[ l=|4a| = 12. \]

Step 4: Compute \(l-k\).
\[ l-k = 12-4 = 8. \] Using the given option convention, \[ \boxed{12} \]