<p>A curve $C$ is defined by the parametric equations</p>
<p style="text-align: center">$x=2{{t}^{2}}$, $y=\sin \left( -t \right)$, for $0\le t\le 3\pi $</p>
<ol class="i-list"><li><div class="list-item-content"><span class="list-item-text">Find the exact coordinates of the axial intercepts and sketch the curve $C$.</span></div><p style="text-align: right;">[3]</p></li><li><div class="list-item-content"><span class="list-item-text">Find the coordinates of the point(s) of intersection between curve $C$ and the line $y=-\frac{1}{2}x+4$.</span></div><p style="text-align: right;">[3]</p></li></ol>

A curve $C$ is defined by the parametric equations

$x=2{{t}^{2}}$, $y=\sin \left( -t \right)$, for $0\le t\le 3\pi $

Find the exact coordinates of the axial intercepts and sketch the curve $C$.

Find the coordinates of the point(s) of intersection between curve $C$ and the line $y=-\frac{1}{2}x+4$.

Parametric Equations

Graphing Techniques

International Baccalaureate Higher Level Mathematics

IB HL Math

ACJC Parametric Equations Tutorial Q4

$y = 0$: $\sin(-t) = 0 \Rightarrow t = 0, \pi, 2\pi, 3\pi$

$x$-intercepts: $(0, 0)$, $(2\pi^{2}, 0)$, $(8\pi^{2}, 0)$, $(18\pi^{2}, 0)$.

Sketch: oscillating curve starting at origin, moving right as $t$ increases.

Sub $x = 2t^{2}$, $y = \sin(-t)$ into $y = -\dfrac{1}{2}x + 4$:

$\sin(-t) = -t^{2} + 4 \Rightarrow t^{2} + \sin(-t) - 4 = 0$

From GC: $t = 2.1932$. $x = 2(2.1932)^{2} = 9.62$, $y = \sin(-2.1932) = -0.812$.

Point of intersection: $(9.62, -0.812)$

ACJC Parametric Equations Tutorial Q4

ACJC Parametric Equations Tutorial Q4

Video Solution: