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

(i)
$\frac{dy}{dx} = \frac{dy}{dt} \times \frac{dt}{dx}$

$\frac{dy}{dx} = - \frac{2}{t^2} \times \frac{1}{2t} = - \frac{1}{t^3}$

At P, $t = p \Rightarrow \frac{dy}{dx} = -\frac{1}{p^3}$

$y - (\frac{2}{p}) = -\frac{1}{p^3} (x - p^2)$

$y = - \frac{x}{p^3} + \frac{3}{p}$

(ii)
At Q, $y = 0, \Rightarrow x = 3p^2$

At P, $x= 0, \Rightarrow y = \frac{3}{p}$

$\therefore~ Q(3p^2,~0) ~\&~ R(0,~\frac{3}{p})$

(iii)
Midpoint $= (\frac{3p^2}{2}, \frac{3}{2p})$

$\Rightarrow x = \frac{3p^2}{2}, y = \frac{3}{2p}$

$p = \frac{3}{2y}$

$\therefore~ 8xy^2 = 27$

KS Comments:

(i) and (ii) are quite direct so long as students do not have any careless mistakes. (iii) tests students on their understanding of parametric equations and the very idea of “p varies”.

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