Horizontal and vertical parametric tangents · source reference

Printed question

A curve $C$ is defined by the parametric equations $x=\theta-\cos\theta$, $y=\theta+\cos\theta$, $0\le\theta\le2\pi$.

(i) Find the coordinate of the point(s) where the tangent is parallel to $x$-axis. [3]

(ii) Find the equation of tangent to the curve $C$ that is parallel to the $y$-axis. [2]

(iii) Find the exact equation of the tangent to the curve at the point $P$ where $\theta=\pi/6$. [3]

(iv) Hence, determine whether the tangent at point $P$ will meet the curve $C$ again. [3]

[modified 2011/SRJC/Prelim/II/3]

Printed working

(i) $dx/d\theta=1+\sin\theta$, $dy/d\theta=1-\sin\theta$, so $dy/dx=\frac{1-\sin\theta}{1+\sin\theta}$.

Tangent parallel to $x$-axis ⇒ $dy/dx=0$ ⇒ $1-\sin\theta=0$ ⇒ $\sin\theta=1$ ⇒ $\theta=\pi/2$ (given interval). Point $(\pi/2-0,\pi/2+0)=(\pi/2,\pi/2)$.

(ii) Tangent parallel to $y$-axis ⇒ $dy/dx$ undefined ⇒ $1+\sin\theta=0$ ⇒ $\sin\theta=-1$ ⇒ $\theta=3\pi/2$. Its $x$ coordinate is $3\pi/2-\cos(3\pi/2)=3\pi/2$, so equation is $x=3\pi/2$.

(iii) At $\theta=\pi/6$, $x=\pi/6-\sqrt3/2$, $y=\pi/6+\sqrt3/2$, and $dy/dx=\frac{1-\sin(\pi/6)}{1+\sin(\pi/6)}=1/3$.

$$y-\left(\frac\pi6+\frac{\sqrt3}2\right)=\frac13\left[x-\left(\frac\pi6-\frac{\sqrt3}2\right)\right],$$ $$y-\frac\pi6-\frac{\sqrt3}2=\frac13x-\frac\pi{18}+\frac{\sqrt3}6\Rightarrow y=\frac13x+\frac\pi9+\frac2{\sqrt3}.$$

(iv) Substitute the parametric equations into the tangent:

$$\theta+\cos\theta=\frac13(\theta-\cos\theta)+\frac\pi9+\frac2{\sqrt3}\Rightarrow\frac23\theta+\frac43\cos\theta=\frac\pi9+\frac2{\sqrt3}.$$

The native intersection sketch has two intersections: $\theta=\pi/6$ and $\theta=\alpha\ne\pi/6$. Thus the tangent intersects $C$ again at parameter $\alpha$.

Printed graph label uses $y=\frac23x+\frac43\cos\theta$ while its horizontal axis is $\theta$; native graph uses the coherent expression $H(\theta)=\frac23\theta+\frac43\cos\theta$, with the original mixed label recorded here.

Teacher notes

The two callout boxes identify (1) the $x$ coordinate of the point where $\theta=3\pi/2$, and (2) the equation of the tangent parallel to the $y$-axis, a vertical line. An $x$ coordinate is not a gradient.

Answer key

(i) $(\pi/2,\pi/2)$; (ii) $x=3\pi/2$; (iii) $y=x/3+\pi/9+2/\sqrt3$; (iv) yes.

Native diagrams