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Exercise 5.5 · Q2

Q.A tunnel through a mountain for a four lane highway is to have a elliptical opening. The total width of the highway (not the opening) is to be 16 m16\,\text m, and the height at the edge of the road must be sufficient for a truck 4 m4\,\text m high to clear if the highest point of the opening is to be 5 m5\,\text m approximately. How wide must the opening be?

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The highest point of the ellipse gives b=5b=5 directly; the truck-clearance requirement gives one more point (8,4)(8,4) that the ellipse must pass through (the road's edge, half of 16 m16\,\text m, at the truck's height); solve for aa.

Step 1. Set up. Centre the ellipse at the road's midpoint, opening x2a2+y2b2=1\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1. Highest point of the opening is 5 m5\,\text m, so b=5⇒b2=25b=5\Rightarrow b^2=25.

Step 2. Use the clearance condition. The road's edge is at x=±8x=\pm8 (half of 16 m16\,\text m); the truck (4 m4\,\text m tall) must clear there, so the ellipse must pass through (8,4)(8,4) (the minimum requirement — the opening is exactly tall enough at the edge).

Step 3. Substitute (8,4)(8,4).

64a2+1625=1⇒64a2=1−1625=925⇒a2=64×259=16009⇒a=403\dfrac{64}{a^2}+\dfrac{16}{25}=1 \Rightarrow \dfrac{64}{a^2}=1-\dfrac{16}{25}=\dfrac9{25} \Rightarrow a^2=\dfrac{64\times25}9=\dfrac{1600}9 \Rightarrow a=\dfrac{40}3.

Step 4. Width of the opening.

2a=803≈26.7 m2a=\dfrac{80}3\approx26.7\,\text m.

✓Final answer

Opening width =2a=803≈26.7 m=2a=\dfrac{80}3\approx26.7\,\text m.

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