Let $P$ be the point on the parabola $y^2 = 3x$ such that $OP$ makes an angle of $\pi/6$ with the $x$-axis, where $O$ is the origin. A normal is drawn to the parabola at $P$, intersecting the axis of the parabola at $Q$. If $S$ is the focus of the parabola, then $SQ$ is equal to
(1) $9$ , (2) $39/4$, (3) $41/4$, (4) $39/2$
MY ATTEMPT:
Let $P = (t^2/3,t)$. Since the slope of $OP$ is $ \frac{1}{\sqrt{3}}$, so $t = 3 \sqrt{3}$.
Now, the equation of the normal to the parabola at $P$ is $$ y=mx-\frac{3}{2}m-\frac{3}{4}m^3. $$ Substitute $y=0$ to get $ x=\frac{3}{2}+\frac{3}{4}m^2 $, now $m = \frac{1}{2\sqrt{3}}$, so $SQ=\frac{13}{16}$.
My answer didn't match with any of the given options. So where am I wrong ?
The m you found is slope of tangent at P, whereas the equation which you wrote requires the value of m as the slope of normal at P.
So $m=-2\sqrt{3}$. Now you should get the correct answer