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