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Miscellaneous Exercise · Q4

Q.An arch is in the form of a semi-ellipse. It is 88 m wide and 22 m high at the centre. Find the height of the arch at a point 1.51.5 m from one end.

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Model the semi-ellipse with centre at the origin, semi-major axis a=4a = 4 m (horizontal) and semi-minor axis b=2b = 2 m (vertical). A point 1.51.5 m from one end corresponds to x=−2.5x = -2.5 m; substitute into the ellipse equation to find the height y≈1.56y \approx 1.56 m.

The problem asks us to find a specific height on a semi-elliptical arch. The natural approach is to set up a coordinate system, write the equation of the ellipse, and then use the given horizontal distance to compute the corresponding vertical coordinate.

An ellipse centred at the origin with horizontal semi-axis aa and vertical semi-axis bb has equation

x2a2+y2b2=1.\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1.

Since the arch is a semi-ellipse (the upper half), we take y≥0y \geq 0.

Setting up the coordinate system

Place the origin at the centre of the base of the arch. The arch is 88 m wide, so it extends from x=−4x = -4 m to x=4x = 4 m. The height at the centre (where x=0x = 0) is 22 m, which is the maximum height of the semi-ellipse.

From this geometry:

  • The semi-major axis (horizontal) is a=4a = 4 m.
  • The semi-minor axis (vertical) is b=2b = 2 m.

The equation of the ellipse becomes

x216+y24=1.\frac{x^2}{16} + \frac{y^2}{4} = 1.

Finding the point of interest

The phrase "a point 1.51.5 m from one end" means 1.51.5 m horizontally from one of the two ends of the arch. Take the left end at x=−4x = -4 m. Moving 1.51.5 m to the right gives

x=−4+1.5=−2.5 m.x = -4 + 1.5 = -2.5 \text{ m}.

(Alternatively, measuring from the right end x=4x = 4 m would give x=4−1.5=2.5x = 4 - 1.5 = 2.5 m, which by symmetry yields the same height.)

Computing the height

Substitute x=−2.5x = -2.5 into the ellipse equation: …

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