Question

A racing track is built around an elliptical ground whose equation is given by \[ 9x^2 + 16y^2 = 144 \] The width of the track is \(3\) m as shown. Based on the given information answer the following: 

(i) Express \(y\) as a function of \(x\) from the given equation of ellipse. 
(ii) Integrate the function obtained in (i) with respect to \(x\). 
(iii)(a) Find the area of the region enclosed within the elliptical ground excluding the track using integration. 
OR 
(iii)(b) Write the coordinates of the points \(P\) and \(Q\) where the outer edge of the track cuts \(x\)-axis and \(y\)-axis in first quadrant and find the area of triangle formed by points \(P,O,Q\). 
 

Hint:

The standard equation of an ellipse is \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\). Its area is \(\pi ab\). Integration of the upper half and multiplying by four gives the full area.

Answer 1

Solution: 

We are given: 
- A racing track built around an elliptical ground with the equation \( 9x^2 + 16y^2 = 144 \). 
- The width of the track is 3 meters. 

(i) Express \( y \) as a function of \( x \) from the given equation of the ellipse. 
The equation of the ellipse is given as: \[ 9x^2 + 16y^2 = 144 \] To express \( y \) as a function of \( x \), solve for \( y \): \[ 16y^2 = 144 - 9x^2 \] \[ y^2 = \frac{144 - 9x^2}{16} \] \[ y = \pm \frac{\sqrt{144 - 9x^2}}{4} \] Thus, \( y \) as a function of \( x \) is: \[ y = \pm \frac{\sqrt{144 - 9x^2}}{4} \] 

(ii) Integrate the function obtained in (i) with respect to \( x \). 
To find the area of the region enclosed by the ellipse (excluding the track), we will integrate the positive part of \( y \) from the equation obtained above: \[ \int_{-a}^{a} y \, dx = \int_{-a}^{a} \frac{\sqrt{144 - 9x^2}}{4} \, dx \] where \( a \) is the half-length of the major axis of the ellipse. From the original equation, we find: \[ \frac{9x^2}{144} = \frac{x^2}{16}, \quad \Rightarrow \quad a = 4 \] Thus, the limits of integration are from \( -4 \) to \( 4 \). Now we proceed to integrate: \[ \int_{-4}^{4} \frac{\sqrt{144 - 9x^2}}{4} \, dx \] This is a standard form integral that can be solved using trigonometric substitution. 

(iii)(a) Find the area of the region enclosed within the elliptical ground excluding the track using integration. 
The area enclosed by the ellipse is found by integrating the entire equation: \[ \text{Area} = 2 \int_0^4 \frac{\sqrt{144 - 9x^2}}{4} \, dx \] 

(iii)(b) Write the coordinates of the points \( P \) and \( Q \) where the outer edge of the track cuts the \( x \)-axis and \( y \)-axis in the first quadrant, and find the area of the triangle formed by points \( P \), \( O \), and \( Q \). 
The points where the outer edge of the track cuts the axes correspond to the points where the major and minor axes of the ellipse intersect the axes. These can be found by substituting \( y = 0 \) and \( x = 0 \) into the ellipse equation. For \( y = 0 \): \[ 9x^2 = 144 \quad \Rightarrow \quad x = \pm 4 \] So, the point \( P \) is at \( (4, 0) \). For \( x = 0 \): \[ 16y^2 = 144 \quad \Rightarrow \quad y = \pm 3 \] So, the point \( Q \) is at \( (0, 3) \). The area of the triangle formed by points \( P(4, 0) \), \( O(0, 0) \), and \( Q(0, 3) \) is: \[ \text{Area of Triangle} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \times 3 = 6 \] 

Final Answers: 
1. \( y = \pm \frac{\sqrt{144 - 9x^2}}{4} \) 
2. The integral to find the area enclosed within the ellipse is: \[ \int_{-4}^{4} \frac{\sqrt{144 - 9x^2}}{4} \, dx \] 
3. The coordinates of the points where the outer edge cuts the axes are: - \( P(4, 0) \) - \( Q(0, 3) \) The area of the triangle formed by \( P(4, 0) \), \( O(0, 0) \), and \( Q(0, 3) \) is \( 6 \) square units.