Graphs and transformations: ACJC 2025 Promo Q5

(a)

The endpoint is at $\theta = 0$, giving $(3, 0)$.

(b)

$$\begin{aligned} \frac{\,\mathrm{d} x}{\,\mathrm{d}\theta} &= 6\sec^2\theta\tan\theta, \qquad \frac{\,\mathrm{d} y}{\,\mathrm{d}\theta} = 2\sec^2\theta\\ \frac{\,\mathrm{d} y}{\,\mathrm{d} x} &= \frac{2\sec^2\theta}{6\sec^2\theta\tan\theta} = \frac{1}{3\tan\theta}. \quad\text{((shown))} \end{aligned}$$

(c)

(i) $L_1$ is parallel to the $y$-axis, so the tangent is vertical and $\tan\theta = 0$, giving $\theta = 0$ and $x = 3\sec^2 0 = 3$. The equation is $x = 3$.

(c)

(ii) At $(3.75, 1)$: $2\tan\theta = 1$, so $\tan\theta = \tfrac{1}{2}$. Hence $$\begin{aligned} \frac{\,\mathrm{d} y}{\,\mathrm{d} x} &= \frac{1}{3\left(\tfrac{1}{2}\right)} = \frac{2}{3}. \end{aligned}$$

(d)

$L_1$ is vertical and $L_2$ has gradient $\tfrac{2}{3}$. Thus $$\begin{aligned} \tan\!\left(\frac{\pi}{2} - \beta\right) = \frac{2}{3} \quad\Rightarrow\quad \tan\beta = \frac{3}{2}. \end{aligned}$$

(e)

Since $y = 2\tan(\pi - \theta) = -2\tan\theta$, the $y$-coordinate is negated while $x$ is unchanged. Hence $C_2$ is a reflection of $C_1$ in the $x$-axis.