Question

If $x+by+c=0$ is normal to the parabola $y^2=12x,$ tben complete set of all values of $c$ is

• A. $(6,\infty )$
• B. $(-\infty ,6)$
• C. $(-\infty ,-6)\cup (6,\infty )$
• D. $(-\infty ,\infty )-\left\{6\right\}$

Answer 1

The correct option is A $(6,\infty )$

$y=mx+c$ is normal to $y^2=4ax$ then $c=-2am-a{m}^3$

$\frac{-c}{b}=2\left(3\right)(-\frac{1}{b})-\left(3\right)(-\frac{1}{b}{)}^3\Rightarrow (6-c)b^2+3=0b^2=\frac{-3}{6-c}\Rightarrow 6-c<0\therefore c>6$