<p>A curve is defined parametrically by $x=\frac{5}{2}\cos \theta ,\,\,y=\frac{3}{2}\sin \theta \,\,,$for $0\le \theta \le 2\pi $.</p>
<ol class="i-list"><li><div class="list-item-paragraph">Find the coordinates of the axial intercepts algebraically.</div><p style="text-align: right;">[2]</p></li><li><div class="list-item-paragraph">Show that the Cartesian equation of the curve is $\frac{{{x}^{2}}}{25}+\frac{{{y}^{2}}}{9}=\frac{1}{4}$.</div><p style="text-align: right;">[2]</p></li><li><div class="list-item-paragraph">Show that the distance of the point $P$ on the curve with parameter $p$ to the point $(2,0)$ is $\frac{5}{2}-2\cos \,p$. Find a similar expression for the distance between $P$ and $(-2,0)$.</div><p style="text-align: right;">[3]</p></li></ol>

A curve is defined parametrically by $x=\frac{5}{2}\cos \theta ,\,\,y=\frac{3}{2}\sin \theta \,\,,$for $0\le \theta \le 2\pi $.

Find the coordinates of the axial intercepts algebraically.

Show that the Cartesian equation of the curve is $\frac{{{x}^{2}}}{25}+\frac{{{y}^{2}}}{9}=\frac{1}{4}$.

Show that the distance of the point $P$ on the curve with parameter $p$ to the point $(2,0)$ is $\frac{5}{2}-2\cos \,p$. Find a similar expression for the distance between $P$ and $(-2,0)$.

Parametric Equations

Graphing Techniques

International Baccalaureate Higher Level Mathematics

IB HL Math

N1995 P2 Q12 [Modified]

To find the $y$-intercepts, let $x = 0 \Rightarrow 0 = \frac{5}{2}\cos\theta$, $\cos\theta = 0$, $\theta = \frac{\pi}{2}, \frac{3\pi}{2}$

$\theta = \frac{\pi}{2}$, $y = \frac{3}{2}\sin\frac{\pi}{2} = \frac{3}{2}$

$\theta = \frac{3\pi}{2}$, $y = \frac{3}{2}\sin\left(\frac{3\pi}{2}\right) = -\frac{3}{2}$

To find the $x$-intercepts, let $y = 0 \Rightarrow \frac{3}{2}\sin\theta = 0$, $\sin\theta = 0 \Rightarrow \theta = 0, \pi$

$\theta = 0$, $x = \frac{5}{2}\cos 0 = \frac{5}{2}$

$\theta = \pi$, $x = \frac{5}{2}\cos(\pi) = -\frac{5}{2}$

$\therefore \left(0, \frac{3}{2}\right), \left(0, -\frac{3}{2}\right), \left(\frac{5}{2}, 0\right), \left(-\frac{5}{2}, 0\right)$

$\cos^2\theta + \sin^2\theta = 1$; $\cos\theta = \frac{2x}{5}$, $\sin\theta = \frac{2y}{3}$

$\left(\frac{2x}{5}\right)^2 + \left(\frac{2y}{3}\right)^2 = 1$

$\frac{4x^2}{25} + \frac{4y^2}{9} = 1$

At point $P$, $\theta = p$
$x = \frac{5}{2}\cos p$, $y = \frac{3}{2}\sin p$

Distance from $P\left(\frac{5}{2}\cos p, \frac{3}{2}\sin p\right)$ to $(2, 0)$

$= \sqrt{\left(\frac{5}{2}\cos p - 2\right)^2 + \left(\frac{3}{2}\sin p - 0\right)^2}$

$= \sqrt{\frac{25}{4}\cos^2 p - 10\cos p + 4 + \frac{9}{4}\sin^2 p}$

$= \sqrt{4\cos^2 p + \frac{9}{4} - 10\cos p + 4}$

$= \sqrt{4\cos^2 p - 10\cos p + \frac{25}{4}}$

$= \sqrt{\frac{16\cos^2 p - 40\cos p + 25}{4}}$

$= \frac{\sqrt{(4\cos p - 5)^2}}{2} = \frac{|4\cos p - 5|}{2}$

Since $4\cos p - 5 < 0$ (as $\cos p \leq 1$), reject $\frac{1}{2}(4\cos p - 5)$.

$= -\frac{1}{2}(4\cos p - 5) = -2\cos p + \frac{5}{2}$ (shown)

Distance from $P$ to $(-2, 0)$

$= \sqrt{\left(\frac{5}{2}\cos p + 2\right)^2 + \left(\frac{3}{2}\sin p\right)^2}$

$= \sqrt{4\cos^2 p + 10\cos p + \frac{25}{4}}$

$= \sqrt{\frac{16\cos^2 p + 40\cos p + 25}{4}} = \frac{\sqrt{(4\cos p + 5)^2}}{2}$

Since $4\cos p + 5 > 0$, $= \frac{1}{2}(4\cos p + 5)$ (shown)

N1995 P2 Q12 [Modified]

N1995 P2 Q12 [Modified]

Video Solution: