Question

The parabola $y^2=4ax$ passes through the centre of the circle $4x^2+4y^2-8x+12y-7=0$. The directrix of the parabola will be

• A. $4x+9=0$
• B. $4x+15=0$
• C. $16x+9=0$
• D. $16x-7=0$

Answer 1

The correct option is C $16x+9=0$

Circle : $4x^2+4y^2-8x+12y-7=0$
$\Rightarrow x^2+y^2-2x+3y-\frac{7}{4}=0$
Centre of this circle $C\equiv (1,-\frac{3}{2})$

Given parabola passes through this centre so,
${(-\frac{3}{2})}^2=4a\left(1\right)$
$\Rightarrow a=\frac{9}{16}$
Directrix : $x=-a$
$x=-\frac{9}{16}\Rightarrow 16x+9=0$