If y=4x+3 is parallel to a tangent to the par | vedclass

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(B) The equation of the parabola is $y^{2}=12x$. Comparing with $y^{2}=4ax$, we get $4a=12$, so $a=3$.The slope of the given line $y=4x+3$ is $m=4$.Th

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$\frac{219}{\sqrt{17}}$

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(B) The equation of the parabola is $y^{2}=12x$. Comparing with $y^{2}=4ax$, we get $4a=12$, so $a=3$.The slope of the given line $y=4x+3$ is $m=4$.The slope of the normal to the parabola $y^{2}=4ax$ at point $(x_{1}, y_{1})$ is given by $m_{n} = -\frac{y_{1}}{2a}$.Since the normal is parallel to the line $y=4x+3$, its slope must be $4$. Thus, $-\frac{y_{1}}{2(3)} = 4$, which gives $y_{1} = -24$.Substituting $y_{1} = -24$ into the parabola equation $y^{2}=12x$, we get $(-24)^{2} = 12x$, so $576 = 12x$, which gives $x_{1} = 48$.The point on the parabola is $(48, -24)$.The equation of the normal at $(48, -24)$ is $y - (-24) = 4(x - 48)$, which simplifies to $y + 24 = 4x - 192$, or $4x - y - 216 = 0$.The given line is $4x - y + 3 = 0$.The distance between two parallel lines $Ax + By + C_{1} = 0$ and $Ax + By + C_{2} = 0$ is given by $d = \frac{|C_{1} - C_{2}|}{\sqrt{A^{2} + B^{2}}}$.Here, $d = \frac{|3 - (-216)|}{\sqrt{4^{2} + (-1)^{2}}} = \frac{|3 + 216|}{\sqrt{16 + 1}} = \frac{219}{\sqrt{17}}$.

If $y=4x+3$ is parallel to a tangent to the parabola $y^{2}=12x$, then its distance from the normal parallel to the given line is

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