All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.



\frac{dy}{dx}={\frac{dy}{dt}}\times {\frac{dt}{dx}}



At P, t=p, \frac{dy}{dx}=\frac{1}{p}
Equation of tangent: y-6p=\frac{1}{p}(x-3p^{2})
At D, x=0,  y=-3p+6p=3p
Therefore, D(0,3p).
Midpoint of PD = (\frac{0+3p^{2}}{2}, \frac{6p+3p}{2})=(\frac{3p^{2}}{2}, \frac{9p}{2})

As p varies, x=\frac{3}{2} p^{2}, y= \frac{9}{2}p
We have that p=\frac{2y}{9}
Then, x=\frac{3}{2}(\frac{2y}{9})^{2}
Thus, 27x=2y^{2}

Personal Comments:
This questions test students on their understanding of parametric equations and finding gradient. Find the cartesian equation can be slightly tricky if a student hasn’t been exposed to such questions before.

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