A curve $C$ has parametric equations $x = 3t^2$, $y = 2t^3$, for $t

Question

A curve $C$ has parametric equations $x = 3t^2$, $y = 2t^3$, for $t < 0$. Sketch $C$.[2] Points $P$ and $Q$ have parameters $p$ and $q$ respectively. Given that $q = 2p$, find the cartesian equation of the locus of the mid-point of $PQ$ as $p$ varies.[2] A curve $L$ has equation $x = y^2 + 1$. It is given that the curves $C$ and $L$ touch at the point $M$. Show that $4t^6 - 3t^2 + 1 = 0$ at $M$. Hence, or otherwise, find the exact coordinates of $M$.[6]

Accepted Answer

$M\left(\dfrac{3}{2}, -\dfrac{1}{\sqrt{2}}\right)$

N2013 P1 Q11 [Modified]

N2013 P1 Q11 [Modified]

Video Solution:

Default solution