Question

The directrix of the the parabola $ 4y^2 + 12x - 12y + 39 = 0 $ is

• A. $ x = \frac{3}{4} $
• B. $ x = \frac{-7}{4} $
• C. $ x = \frac{-5}{2} $
• D. $ x = \frac{3}{2} $

Answer 1

Answer: (B)

We have,
$4y^2 + 12 x - 12 y + 39 = 0$
$\Rightarrow [(2y)^2 - 2\times 2y \times 3 + 9] - 9 + 12 x + 39 = 0$
$\Rightarrow (2y - 3)^2 + 12x + 30 = 0$
$\Rightarrow (2y -3)^2 = -12x - 30$
$ \Rightarrow 4(y - \frac{3}{2})^2 = -12 (x + \frac{5}{2})$
On comparing this equation with
$Y_n^2 = - 4aX$, we get
$a = \frac{3}{4}$
Now, equation of directrix is $X = a$
$\therefore x + \frac{5}{2} = \frac{3}{4}$
$\Rightarrow x = \frac{3}{4} - \frac{5}{2} $
$\Rightarrow x = \frac{3 - 10}{4}$
$\Rightarrow x = - \frac{7}{4}$