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

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(A) The equation of the line is $x + by + c = 0$,which can be rewritten as $y = -\frac{1}{b}x - \frac{c}{b}$.Comparing this with the slope-intercept f

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$(-\infty, -6)$

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(A) The equation of the line is $x + by + c = 0$,which can be rewritten as $y = -\frac{1}{b}x - \frac{c}{b}$.Comparing this with the slope-intercept form $y = mx + k$,we get $m = -\frac{1}{b}$ and $k = -\frac{c}{b}$.The condition for the line $y = mx + k$ to be a normal to the parabola $y^2 = 4ax$ is $k = -2am - am^3$.Here,$4a = 12$,so $a = 3$.Substituting the values,we get $-\frac{c}{b} = -2(3)(-\frac{1}{b}) - 3(-\frac{1}{b})^3$.$-\frac{c}{b} = \frac{6}{b} + \frac{3}{b^3}$.Multiplying by $b^3$ (assuming $b eq 0$),we get $-cb^2 = 6b^2 + 3$,which simplifies to $b^2(6 + c) = -3$.Thus,$b^2 = -\frac{3}{6 + c}$.Since $b^2 > 0$ for any real $b eq 0$,we must have $-\frac{3}{6 + c} > 0$.This implies $6 + c < 0$,so $c < -6$.

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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