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 written as $y = -\frac{1}{b}x - \frac{c}{b}$.Comparing this with $y = mx + k$,we get $m

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$(-\infty, -6)$

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