A racing track is build around an elliptical ground whose equation is given by \(9x^2 + 16y^2 = 144\). The width of the track is \(3\text{ m}\). Based on the given information, answer the following question:
Question: 1
Express \(y\) as a function of \(x\) from the given equation of ellipse.
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When isolating variables from the equation of a conic section like an ellipse \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\):
- Standard form can be obtained first by dividing by the constant: \(\frac{9x^2}{144} + \frac{16y^2}{144} = 1 \Rightarrow \frac{x^2}{16} + \frac{y^2}{9} = 1\).
- Shifting terms gives \(\frac{y^2}{9} = 1 - \frac{x^2}{16}\).
- Multiplying by \(9\) and pulling out the common denominator yields \(y^2 = \frac{9}{16}(16-x^2)\). Always remember that taking a square root on an even power produces both positive and negative branches (\(\pm\)), representing the upper and lower halves of the ellipse respectively!
Integrate the function obtained in (i) with respect to \(x\).
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Always remember the structure of the special standard integral formula: \(\int \sqrt{a^2 - x^2} dx = \frac{x}{2}\text{(the radical)} + \frac{a^2}{2}\sin^{-1}(\frac{x}{a})\). It acts as a major building block whenever computing total areas under circles or ellipses via integration techniques!
Find the area of the region enclosed within the elliptical ground excluding the track using integration.
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The standard formula for the complete area of an ellipse \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) is simply \(\text{Area} = \pi a b\). Here, \(a = 4\) and \(b = 3\), so \(\text{Area} = \pi \times 4 \times 3 = 12\pi\). Use this short cross-verification method in exams to guarantee absolute accuracy!
Write the co-ordinates of the points P and Q where the outer edge of the track cuts \(x\)-axis and \(y\)-axis in the first quadrant and find the area of the triangle formed by points P, O, Q using integration.
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For a right-angled triangle formed by the origin and the axes intercepts, you can quickly double-check your integration answer using the elementary geometry formula: \(\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}\). Here, the base is \(\text{OP} = 7\) and height is \(\text{OQ} = 6\), giving \(\text{Area} = \frac{1}{2} \times 7 \times 6 = 21\). It provides instant verification!
