From the focus of the parabola y^2 = 12x, a r | vedclass

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(B) The parabola is $y^2 = 12x$, so $4a = 12$, which implies $a = 3$. The focus is $(3, 0)$.The incident ray passes through the focus $(3, 0)$ with sl

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$y = 18$

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(B) The parabola is $y^2 = 12x$, so $4a = 12$, which implies $a = 3$. The focus is $(3, 0)$.The incident ray passes through the focus $(3, 0)$ with slope $m = 	an(	an^{-1} \frac{3}{4}) = \frac{3}{4}$.The equation of the incident ray is $y - 0 = \frac{3}{4}(x - 3)$, or $x = \frac{4y}{3} + 3$.Substituting this into the parabola equation $y^2 = 12x$:$y^2 = 12(\frac{4y}{3} + 3) = 16y + 36$.$y^2 - 16y - 36 = 0$.$(y - 18)(y + 2) = 0$.Since the ray is directed into the first quadrant, we take $y = 18$.$A$ property of the parabola states that any ray passing through the focus reflects parallel to the axis of the parabola.The axis of the parabola $y^2 = 12x$ is the $x$-axis $(y = 0)$.Thus, the reflected ray is a horizontal line passing through the point $P$ with $y$-coordinate $18$.The equation of the reflected ray is $y = 18$.

From the focus of the parabola $y^2 = 12x$, a ray of light is directed in a direction making an angle $\tan^{-1} \frac{3}{4}$ with the $x$-axis. Then the equation of the line along which the reflected ray leaves the parabola is

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