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

Step 1: Convert the ellipse into standard form.
Given equation \[ 9x^2 + 16y^2 = 144 \] Divide the entire equation by \(144\): \[ \frac{x^2}{16} + \frac{y^2}{9} = 1 \] Thus the ellipse has \[ a = 4, \quad b = 3 \] Step 2: Express \(y\) as a function of \(x\).
From the equation \[ \frac{x^2}{16} + \frac{y^2}{9} = 1 \] \[ \frac{y^2}{9} = 1 - \frac{x^2}{16} \] \[ y^2 = 9\left(1 - \frac{x^2}{16}\right) \] \[ y = 3\sqrt{1 - \frac{x^2}{16}} \] Thus \[ y = 3\sqrt{1 - \frac{x^2}{16}} \] Step 3: Integrate the function.
We integrate \[ \int y \, dx \] Substitute \(y\): \[ \int 3\sqrt{1 - \frac{x^2}{16}} \, dx \] \[ = 3\int \sqrt{1 - \frac{x^2}{16}}\, dx \] Let \[ x = 4\sin\theta \] Then \[ dx = 4\cos\theta\, d\theta \] Also \[ \sqrt{1-\frac{x^2}{16}}=\sqrt{1-\sin^2\theta}=\cos\theta \] Thus \[ 3\int \cos\theta \cdot 4\cos\theta\, d\theta \] \[ =12\int \cos^2\theta\, d\theta \] Using identity \[ \cos^2\theta=\frac{1+\cos2\theta}{2} \] \[ =12\int \frac{1+\cos2\theta}{2} d\theta \] \[ =6(\theta+\frac{\sin2\theta}{2})+C \] Step 4: Area enclosed by the ellipse.
Area of ellipse using integration \[ A = 4\int_0^4 y\,dx \] Substitute \(y\): \[ A = 4\int_0^4 3\sqrt{1-\frac{x^2}{16}}dx \] \[ A = 12\int_0^4 \sqrt{1-\frac{x^2}{16}}dx \] Using standard result \[ \int_0^a \sqrt{1-\frac{x^2}{a^2}}dx=\frac{\pi a}{4} \] Thus \[ \int_0^4 \sqrt{1-\frac{x^2}{16}}dx = \pi \] Therefore \[ A = 12\pi \] Hence area of ellipse \[ A = \pi ab \] \[ = \pi (4)(3) \] \[ = 12\pi \] Step 5: Coordinates of intercepts of outer track.
Since track width = \(3\) m, the semi axes increase by \(3\).
Thus outer ellipse has \[ a = 7, \quad b = 6 \] Thus intercepts are \[ P(7,0), \quad Q(0,6) \] Step 6: Area of triangle \(POQ\).
Area of triangle with intercepts: \[ A = \frac12 \times 7 \times 6 \] \[ A = 21 \] Final Answer:
\[ y = 3\sqrt{1 - \frac{x^2}{16}} \] Area of ellipse: \[ \boxed{12\pi} \] Coordinates \[ P(7,0), \quad Q(0,6) \] Area of triangle \(POQ\): \[ \boxed{21} \]