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Original question & working

Fit an elliptical tunnel from boundary points · source reference

2008/ACJC/Prelim/II/2 · Question pp. 4 · Working pp. 12

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The working and teaching notes are included in the interactive lesson.

Source qualification: solution PDF p.12 assumes exact contact at the load corners. Its margin notes use metres, but the question dimensions are centimetres.

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Full original question, working & notes

Printed question

The diagram below shows the cross-section of a single track tunnel entrance for steam-engine in a model train system. The elliptical entrance of the tunnel can be modelled by the equation x2+Ay2+By+C=0x^2+Ay^2+By+C=0.

The highest point of the tunnel entrance is 7 cm directly above the origin at ground level. The track, mounted on track bed PSPS, is centred about the origin. A train carrying containers which measured 2 cm wide may stack goods as high as 6.8 cm.

Solve for the constants A,B,CA,B,C, to 4 decimal places. Hence find the length of the widest part of the tunnel, to the nearest 0.1 cm. [6]

Native tunnel and train diagram: 7 cm high tunnel, 4.3 cm track bed PS, 2 cm wide and 6.8 cm high container

[2008/ACJC/Prelim/II/2]

Printed working

x2+Ay2+By+C=0x^2+Ay^2+By+C=0.

Sub (2.15,0)(2.15,0), C=−2.152=−4.6225C=-2.15^2=-4.6225.

Sub (1,6.8)(1,6.8), 1+6.82A+6.8B+C=01+6.8^2A+6.8B+C=0.

46.24A+6.8B=3.622546.24A+6.8B=3.6225.

Sub (0,7)(0,7), 49A+7B+C=049A+7B+C=0.

49A+7B=4.622549A+7B=4.6225.

Solving, A=0.6381827731=0.6382A=0.6381827731=0.6382 (4 d.p.),

B=−3.806922269=−3.8069B=-3.806922269=-3.8069 (4 d.p.).

x2+0.63818y2−3.8069y−4.6225=0x^2+0.63818y^2-3.8069y-4.6225=0.

x2+0.63818(y2−5.9653y)−4.6225=0x^2+0.63818(y^2-5.9653y)-4.6225=0.

x2+0.63818[(y−2.9826)2−8.8961]−4.6225=0x^2+0.63818[(y-2.9826)^2-8.8961]-4.6225=0.

x2+0.63818(y−2.9826)2=10.300x^2+0.63818(y-2.9826)^2=10.300.

When y=2.9826y=2.9826, x2=10.300∴x=3.2093x^2=10.300\therefore x=3.2093.

Widest part of tunnel is 6.4 cm (to 1 d.p.).

Teacher notes

For train carrying containers with goods as high as 6.8m to pass through tunnel, the train must be centred at the origin (i.e. 1m on either side of origin along OxOx direction).

For ellipse, (x−h)2a2+(y−k)2b2=1\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1, the widest part along xx-direction is ∣a∣|a| and it occurs when (y−k)2=0(y-k)^2=0 (i.e. along line of symmetry y=ky=k).

Printed answer key

A=0.6382,B=−3.8069,C=−4.6225A=0.6382,B=-3.8069,C=-4.6225 (4 d.p.); 6.4 cm.

Editorial qualification

The teacher note prints “6.8m” and “1m”; the question and diagram use centimetres, so these are unit typos. Also ∣a∣|a| is the semi-width of the ellipse; the full widest length is 2∣a∣2|a|, which the printed final 6.4 cm uses correctly.

Source provenance

2025 TJC H2 Math (9758) JC2 MSM: EQUATIONS. Question sheet printed pages 15–19; solution sheet printed pages 1–12. Original exam attribution is retained above. Currency dollar signs are spelled out outside maths to avoid confusing currency with TeX delimiters.