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: Convert the parabola into standard form.
Given parabola: \[ x^2+4y-6x+\lambda=0 \] Rearranging, \[ x^2-6x=-4y-\lambda \] Completing the square, \[ x^2-6x+9=-4y-\lambda+9 \] \[ (x-3)^2=-4y-\lambda+9 \] \[ (x-3)^2=-4\left(y+\frac{\lambda-9}{4}\right) \] Comparing with the standard form \[ (x-h)^2=4a(y-k), \] we get \[ h=3,\qquad k=-\frac{\lambda-9}{4},\qquad 4a=-4. \] Hence, \[ a=-1. \]

Step 2: Use the equation of the directrix.
For the parabola \[ (x-h)^2=4a(y-k), \] the directrix is \[ y=k-a. \] Since \[ a=-1, \] the directrix becomes \[ y=k+1. \] Given directrix: \[ y+1=0 \] or \[ y=-1. \] Therefore, \[ k+1=-1 \] \[ k=-2. \]

Step 3: Find the value of \(\lambda\).
We know \[ k=-\frac{\lambda-9}{4}. \] Substituting \(k=-2\), \[ -\frac{\lambda-9}{4}=-2. \] Multiplying by \(4\), \[ -(\lambda-9)=-8. \] \[ -\lambda+9=-8. \] \[ -\lambda=-17. \] \[ \lambda=17. \] Thus, options (1) and (2) are incorrect.

Step 4: Find the focus.
For the parabola \[ (x-h)^2=4a(y-k), \] the focus is \[ (h,\;k+a). \] Substituting \[ h=3,\quad k=-2,\quad a=-1, \] we get \[ (3,\,-2-1) \] \[ (3,-3). \] Hence, the focus is \[ (3,-3). \]

Step 5: Check the vertex.
The vertex is \[ (h,k) = (3,-2). \] Therefore, \[ (3,-3) \] is not the vertex.
Thus, option (4) is incorrect.

Step 6: Final conclusion.
The correct statement is \[ \boxed{\text{Focus is }(3,-3)}. \]