Question 66 of 70intermediate🔧 ApplyLong Answer4 marks

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.

Correct Answer

The height of the arch at a point 1.51.5 m from one end is approximately 1.561.56 m.

Exercise: Miscellaneous Exercise on Chapter 10 | Q: 4 | (Page 29)
For More Understanding

Explanation

This is a direct application of the ellipse equation. The arch forms a semi-ellipse with the major axis along the horizontal direction. The width of 88 m gives the major axis length (2a2a), and the height of 22 m at centre gives the semi-minor axis (bb).

Using the standard ellipse equation, we substitute the xx-coordinate (found by measuring 1.51.5 m from one end) to find the corresponding yy-value, which represents the height.

Solution Steps

  1. Step 1: Identify the parameters of the ellipse. The arch is 88 m wide, so the major axis length is 88 m. Therefore, the semi-major axis a=4a = 4 m. The height at the centre is 22 m, so the semi-minor axis b=2b = 2 m.

  2. Step 2: Set up the coordinate system with the centre of the ellipse at the origin. The equation of the ellipse is:

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

    which gives:

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

  3. Step 3: Find the xx-coordinate of the point 1.51.5 m from one end. The ends of the arch are at x=−4x = -4 m and x=4x = 4 m. A point 1.51.5 m from one end (say from x=−4x = -4) is at x=−4+1.5=−2.5x = -4 + 1.5 = -2.5 m (or equivalently x=2.5x = 2.5 m by symmetry).

  4. Step 4: Substitute x=2.5x = 2.5 in the ellipse equation:

    (2.5)216+y24=1\frac{(2.5)^2}{16} + \frac{y^2}{4} = 1

    This gives:

    6.2516+y24=1\frac{6.25}{16} + \frac{y^2}{4} = 1

  5. Step 5: Solve for yy:

    y24=1−6.2516=9.7516\frac{y^2}{4} = 1 - \frac{6.25}{16} = \frac{9.75}{16}

    Therefore, y2=3916y^2 = \frac{39}{16}, and:

    y=394≈1.56 my = \frac{\sqrt{39}}{4} \approx 1.56 \text{ m}
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