Specialist Maths — Question 14

QCAA 2024, Paper 1 · 5 marks

Q14 · 2024 · Technology-freeSimple familiar5 marks

QUESTION 14 (5 marks)

The displacement (cm) of a particle from the origin as it travels in two-dimensional space at time

t

for

0\le t<\frac{\pi}{2}

seconds is given by

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

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)

\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1,

where

a,b\ne0

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

[3 marks]

Determine the centre of the hyperbolic path of the particle.

[1 mark]

WORKED SOLUTION

5 marks

ANSWER

x=2\sec t-1

y=\tan t

;

\frac{(x+1)^2}{4}-y^2=1

; centre

(-1,0)

Worked solution

x=2\sec t-1

y=\tan t

b) From

x=2\sec t-1

\sec t=(x+1)/2

. Using

\tan^2t+1=\sec^2t

and

y=\tan t

y^2+1=\left(\frac{x+1}{2}\right)^2,

\frac{(x+1)^2}{4}-y^2=1.

c) The centre is

(-1,0)

correctly expresses the path as a pair of parametric equations

[1 mark]

expresses

\sec(t)

as the subject of the formula

[1 mark]

uses a suitable Pythagorean identity

[1 mark]

expresses the path in general Cartesian form

[1 mark]

determines the centre of the path

[1 mark]

One QCAA sample method typeset for web; criterion wording is adapted from the official marking guide.

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