If x + by + c = 0 is a normal to the parabola | vedclass

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(A) The equation of the normal to the parabola $y^2 = 4ax$ with slope $m$ is $y = mx - 2am - am^3$.Given the parabola $y^2 = 12x$,we have $4a = 12$,so

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$(-\infty, -6)$

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(A) The equation of the normal to the parabola $y^2 = 4ax$ with slope $m$ is $y = mx - 2am - am^3$.Given the parabola $y^2 = 12x$,we have $4a = 12$,so $a = 3$.The given line is $x + by + c = 0$,which can be rewritten as $y = -\frac{1}{b}x - \frac{c}{b}$.Comparing this with $y = mx + c'$,we get $m = -\frac{1}{b}$ and $c' = -\frac{c}{b}$.Substituting these into the normal equation $c' = -2am - am^3$:$-\frac{c}{b} = -2(3)(-\frac{1}{b}) - 3(-\frac{1}{b})^3$$-\frac{c}{b} = \frac{6}{b} + \frac{3}{b^3}$Multiplying by $-b$:$c = -6 - \frac{3}{b^2}$Since $b^2 > 0$,we have $\frac{3}{b^2} > 0$,which implies $-\frac{3}{b^2} Therefore,$c = -6 - \frac{3}{b^2} Thus,the set of all values of $c$ is $(-\infty, -6)$.

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

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