Question

If the directrix of the parabola \[ x^2+4y-6x+\lambda=0 \] is \(y+1=0\), then which of the following is correct?

• A. \(\lambda=-17\)
• B. \(\lambda=-19\)
• C. Focus is \((3,-3)\)
• D. Vertex is \((3,-3)\)

Hint:

For the parabola \[ (x-h)^2=4a(y-k), \] remember: \[ \text{Vertex }=(h,k), \] \[ \text{Focus }=(h,k+a), \] and \[ \text{Directrix } y=k-a. \]

Answer 1

The Correct Option is C

Step 1: Complete the square.
\(x^2+4y-6x+\lambda=0 \Rightarrow (x-3)^2 = -4y+9-\lambda = -4\left(y-\frac{9-\lambda}{4}\right)\). Vertex: \(\left(3,\frac{9-\lambda}{4}\right)\), opens downward, \(p=1\).

Step 2: Use the directrix condition.
For \((x-h)^2=-4p(y-k)\), directrix is \(y = k+p\). So \(\frac{9-\lambda}{4}+1 = -1 \Rightarrow \frac{9-\lambda}{4} = -2 \Rightarrow \lambda = 17\).

Step 3: Find the focus.
Vertex \(= (3,-2)\), focus at \(y = k-p = -2-1 = -3\). Focus \(= (3,-3)\).
\[ \boxed{\text{Focus is } (3,-3)} \]