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Step-by-Step Solution
Step 1: Set up coordinates for point P on the parabola
Let the parabola be given by
$y^2 = 12x.$
Suppose $P$ has coordinates $(x_1, y_1)$ on this parabola. From the equation of the parabola, we have
$x_1 = \frac{y_1^2}{12}.$
Step 2: Coordinates of N, the foot of the perpendicular from P to the axis
The axis of the parabola $y^2 = 12x$ is the $x$-axis. So, the foot of the perpendicular from
$P(x_1, y_1)$ to the $x$-axis has the same $x$-coordinate and $y=0$. Thus
$N(x_1, 0).$
Step 3: Coordinates of M, the midpoint of PN
The midpoint $M$ of the segment $PN$, where
$P(x_1, y_1)$ and $N(x_1, 0)$, is
$M\bigl(\,x_1,\;\tfrac{y_1}{2}\bigr).$
Step 4: Coordinates of Q, where the line through M parallel to the axis meets the parabola
A line parallel to the axis (the $x$-axis) through
$M\bigl(x_1,\tfrac{y_1}{2}\bigr)$ is a horizontal line
$y = \tfrac{y_1}{2}.$
To find $Q$, we note that $Q$ also lies on the parabola
$y^2 = 12x.$
Substituting $y = \tfrac{y_1}{2}$ into the parabola’s equation:
$
\left(\tfrac{y_1}{2}\right)^2 = 12x
\quad \Longrightarrow \quad
\tfrac{y_1^2}{4} = 12x
\quad \Longrightarrow \quad
x = \frac{y_1^2}{48}.
$
Thus,
$
Q\Bigl( \tfrac{y_1^2}{48}, \;\tfrac{y_1}{2}\Bigr).
$
Step 5: Equation of the line NQ and its y-intercept
The points on this line are
$N(x_1, 0)\, \bigl(x_1 = \tfrac{y_1^2}{12}\bigr)$ and
$Q\Bigl( \tfrac{y_1^2}{48}, \;\tfrac{y_1}{2}\Bigr).$
Let us find the slope $m$ of $NQ$:
$
m
= \frac{\tfrac{y_1}{2} - 0}{\tfrac{y_1^2}{48} - \tfrac{y_1^2}{12}}.
$
Notice
$
\tfrac{y_1^2}{12} = \tfrac{4y_1^2}{48}.
$
So,
$
\tfrac{y_1^2}{48} - \tfrac{y_1^2}{12}
= \tfrac{y_1^2}{48} - \tfrac{4y_1^2}{48}
= -\tfrac{3y_1^2}{48}.
$
Hence
$
m
= \frac{\tfrac{y_1}{2}}{-\tfrac{3y_1^2}{48}}
= \frac{\tfrac{y_1}{2}}{1}\,\cdot \frac{-48}{3y_1^2}
= -\frac{8}{y_1}.
$
The line passing through $N(x_1, 0)$ with slope $m$ is
$
y - 0 = m (x - x_1).
$
Substituting $m = -\tfrac{8}{y_1}$ and $x_1 = \tfrac{y_1^2}{12}$, we get
$
y = -\frac{8}{y_1}\,\bigl(x - \tfrac{y_1^2}{12}\bigr).
$
To find the $y$-intercept, set $x = 0$:
$
y_{\text{intercept}}
= -\frac{8}{y_1}\,\bigl(0 - \tfrac{y_1^2}{12}\bigr)
= \frac{8}{y_1}\,\cdot \frac{y_1^2}{12}
= \frac{8 y_1^2}{12 y_1}
= \frac{2y_1}{3}.
$
Step 6: Use the given y-intercept to find $y_1$
It is given that the $y$-intercept of $NQ$ is
$\frac{4}{3}.$
Hence,
$
\frac{2y_1}{3} = \frac{4}{3}.
$
Solving,
$
2y_1 = 4
\quad \Longrightarrow \quad
y_1 = 2.
$
Step 7: Compute MQ
Now that $y_1 = 2$, let us find $M$ and $Q$ explicitly:
$
x_1 = \frac{y_1^2}{12} = \frac{4}{12} = \frac{1}{3},
\quad
M\Bigl(\frac{1}{3},1\Bigr),
\quad
Q\Bigl(\frac{y_1^2}{48}, \;\frac{y_1}{2}\Bigr)
= \Bigl(\frac{4}{48},1\Bigr)
= \Bigl(\frac{1}{12},1\Bigr).
$
The distance $MQ$ is purely horizontal (since both have $y=1$):
$
MQ
= \left| \frac{1}{3} - \frac{1}{12} \right|
= \frac{4}{12} - \frac{1}{12}
= \frac{3}{12}
= \frac{1}{4}.
$
Final Answer
$\displaystyle MQ = \frac{1}{4},$ which confirms the correct option.
