Specialist Mathematics · Past QCAA questions All simple familiar questions
2024 · Paper 1 · Simple familiar

Question 14

[5 marks] Technology-free
Topic guidePremiumUnit 3 Topic 4.1 · Converting to a Cartesian equation
Game planPremium4.1 Naming and converting a path
The question

The displacement (cm) of a particle from the origin as it travels in two-dimensional space at time $t$ seconds is given by

$$\mathbf{r}(t)=\big(2\sec(t)-1\big)\hat{\mathbf{i}}+\tan(t)\,\hat{\mathbf{j}}$$

a)

Express the path of the particle as a pair of parametric equations.

[1 mark]

A general Cartesian form of a hyperbola with centre $(h,k)$ is $\dfrac{(x-h)^2}{a^2}-\dfrac{(y-k)^2}{b^2}=1$, where $a,b\gt0$.

b)

Use a suitable Pythagorean identity to show that the path of the particle can be expressed in this general Cartesian form.

[3 marks] c)

Determine the centre of the hyperbolic path of the particle.

[1 mark]
Watch the situation first

Secant and tangent trace out a hyperbola

Watch the particle trace its path, then see the whole hyperbola it belongs to and where its centre is.

At t = 0 the particle is at (1, 0) As t increases, it traces a curve sec and tan together give a hyperbola The centre is halfway between the two branches −21412−1xy (−1, 0) x = 2 sec t − 1 y = tan t tan² t + 1 = sec² t y² + 1 = ((x + 1) ÷ 2)² (x + 1)² ÷ 4 − y² = 1

The particle only travels along part of the hyperbola. The centre is where the dashed asymptotes cross.

Now for the mathematics

Move through time and check the Cartesian equation

Move the slider to change $t$. At every time, the particle’s coordinates make $\frac{(x+1)^2}{4}-y^2$ equal to 1.

−4−22442−2−4xy (−1, 0)

Since $\sec t\ge1$ when $\cos t\gt0$, the particle stays on the right branch, where $x\ge1$. The dashed left branch is part of the same hyperbola, but the particle doesn’t visit it here.

QCAA marking guide · 5 marks

Work through the solution one mark at a time

Try each step yourself before you reveal it.

Step 1 · part a)1 mark

Read the two components of the vector.

$$x=2\sec(t)-1\qquad y=\tan(t)$$

Marker: correctly expresses the path of the particle as a pair of parametric equations

Step 2 · part b)1 mark

Make $\sec(t)$ the subject of the $x$ equation, so it can go into an identity.

$$\sec(t)=\frac{x+1}{2}$$

Marker: expresses $\sec(t)$ as the subject of the formula

Step 3 · part b)1 mark

Secant and tangent are linked by this Pythagorean identity.

$$\tan^2(A)+1=\sec^2(A)$$

Using $\sin^2(A)+\cos^2(A)=1$ is also accepted.

Marker: uses a suitable Pythagorean identity

Step 4 · part b)1 mark

Substitute $\tan(t)=y$ and $\sec(t)=\frac{x+1}{2}$, then rearrange into the general form.

$$\begin{aligned}y^2+1&=\left(\frac{x+1}{2}\right)^2\\\frac{(x+1)^2}{4}-y^2&=1\end{aligned}$$

This matches the general form with $h=-1$, $k=0$, $a=2$ and $b=1$.

Marker: expresses path of the particle in general Cartesian form

Step 5 · part c)1 mark

Read $(h,k)$ from the general form. Watch the sign: $(x+1)^2$ is $(x-(-1))^2$.

Centre $(-1,\ 0)$

Marker: determines the centre of the path of the particle

Putting it all together

Each phrase of the question gave you something

“a pair of parametric equations”
One equation for $x$ and one for $y$, both in terms of $t$.
“Use a suitable Pythagorean identity”
Secant and tangent point to $\tan^2+1=\sec^2$.
“show that … in this general Cartesian form”
The form is given, so the marks are for the steps that reach it.
“Determine the centre”
Read $(h,k)$ from your equation, taking care with the signs.
What makes this simple familiar

Secant and tangent in the two components always mean a hyperbola, and the question names both the identity and the form to reach.

If I had to guess where you might lose marks, it would be part c). The centre is $(-1,0)$, not $(1,0)$, because $(x+1)^2$ means $h=-1$. In part b), square the whole of $\frac{x+1}{2}$, including the 2.

Only mark this done when you could do it without help. Reading the solution does not count.

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Question wording and marking-guide steps are from the 2024 QCAA Specialist Mathematics external assessment, © State of Queensland (QCAA) 2024, licensed under CC BY 4.0, and have been adapted. Tangent Tuition is not affiliated with the QCAA.

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