Question

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

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

Answer 1

Answer: (C)

$16x+9=0$

Given circle is $4x^2+4y^2-8x+12y-7=0$
$\Rightarrow x^2+y^2-2x+3x-\frac{7}{4}=0$
Thus centre of the circle is $C(1,-3/2)$
It is given that the parabola $y^2=2ax$ is passing through
centre of the circle
$\Rightarrow {\left(\frac{-3}{2}\right)}^2=2a\times 1\Rightarrow a=\frac{9}{8}$
Hence the directrix of the parabola $y^2=2ax$ is,
$x=-\frac{1}{2}a=-\frac{9}{16}\Rightarrow 16x+9=0$
Hence option 'C' is correct choice.