To express y as a function of x, we need to isolate y from the given equation of the ellipse. We move the x² term to the right side, then divide by 16, and finally take the square root. We factor out 9 from the term inside the square root to simplify further.

To express y as a function of x, we need to isolate y from the given equation of the ellipse. We move the x² term to the right side, then divide by 16, and finally take the square root. We factor out 9 from the term inside the square root to simplify further.

We are asked to integrate the function obtained in part (i) with respect to x. We will use the positive part of the function, y = (3)/(4)√(16 - x²). This integral involves the standard formula for ∫ √(a² - x²) dx. Here, a=4, so we substitute this value into the formula and include the constant factor (3)/(4).

We are asked to integrate the function obtained in part (i) with respect to x. We will use the positive part of the function, y = (3)/(4)√(16 - x²). This integral involves the standard formula for ∫ √(a² - x²) dx. Here, a=4, so we substitute this value into the formula and include the constant factor (3)/(4).

Before calculating the area, we first determine the semi-major and semi-minor axes of the original elliptical ground by converting its equation to the standard form (x²)/(a²) + (y²)/(b²) = 1. The semi-major axis a is the length along the x -axis from the center, and the semi-minor axis b is the length along the y -axis from the center.

Before calculating the area, we first determine the semi-major and semi-minor axes of the original elliptical ground by converting its equation to the standard form (x²)/(a²) + (y²)/(b²) = 1. The semi-major axis a is the length along the x -axis from the center, and the semi-minor axis b is the length along the y -axis from the center.

38. A racing track is build around an elliptical ground whose equation is given by 9x² + 16y² = 144. The width of the track is 3 m as shown below : Based on given information, answer the following questions : (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 co-ordinates 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 the triangle formed by points P, O, Q using integration. (ellipse, equation, integration, area, coordinates)

Answer: (i) $y = \pm\frac{3}{4}\sqrt{16 - x^2}$ (ii) $\frac{3}{4}\left[\frac{x}{2}\sqrt{16 - x^2} + 8\sin^{-1}\left(\frac{x}{4}\right)\right] + C$ (iii) (a) The area of the region enclosed within the elliptical ground excluding the track is $12\pi \text{ m}^2$ . (iii) (b) The coordinates of P are $(5.5, 0)$ and Q are $(0, 4.5)$ . The area of the triangle POQ is $12.375 \text{ m}^2$ .

Step-by-step solution

Step 1: Express y as a function of x

To express $y$ as a function of $x$ , we need to isolate $y$ from the given equation of the ellipse. We move the $x^2$ term to the right side, then divide by 16, and finally take the square root. We factor out 9 from the term inside the square root to simplify further.

Step 2: Integrate the positive function of y with respect to x

We are asked to integrate the function obtained in part (i) with respect to $x$ . We will use the positive part of the function, $y = \frac{3}{4}\sqrt{16 - x^2}$ . This integral involves the standard formula for $\int \sqrt{a^2 - x^2} \, dx$ . Here, $a=4$ , so we substitute this value into the formula and include the constant factor $\frac{3}{4}$ .

Step 3: Find semi-major and semi-minor axes of the inner ellipse

Before calculating the area, we first determine the semi-major and semi-minor axes of the original elliptical ground by converting its equation to the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ . The semi-major axis $a$ is the length along the $x$ -axis from the center, and the semi-minor axis $b$ is the length along the $y$ -axis from the center.

Step 4: Calculate dimensions of the outer ellipse

The width of the track is 3 m. This means the track is 3 m wide, effectively adding 1.5 m to each semi-axis length. So, we add 1.5 m to the semi-major and semi-minor axes of the inner ellipse to find the corresponding values for the outer ellipse.

Step 5: Calculate area of the inner ellipse

The area of an ellipse is given by the formula $\pi ab$ , where $a$ and $b$ are the semi-major and semi-minor axes, respectively. We use the dimensions of the inner ellipse to calculate its area.

Step 6: Calculate area of the outer ellipse

Similarly, we calculate the area of the outer ellipse using its semi-major and semi-minor axes. These dimensions were determined by adding the track width to the inner ellipse's dimensions.

Step 7: Calculate area excluding the track (Area of the track)

The area of the region enclosed within the elliptical ground excluding the track is the area of the inner ellipse, as the question asks for the area of the elliptical ground *excluding* the track. The track is the region *between* the outer and inner ellipses. If the question meant the *area of the track*, it would be the difference between the outer and inner ellipse areas. However, based on the phrasing "area of the region enclosed within the elliptical ground excluding the track", it refers to the area of the inner ellipse. Let's re-read carefully: "area of the region enclosed within the elliptical ground excluding the track". This means the area of the inner ellipse. This interpretation is consistent with typical problem structures where 'ground' is the inner boundary.

Step 8: Final answer for part (iii) (a)

The area of the region enclosed within the elliptical ground, excluding the track, is simply the area of the inner ellipse. This is $12\pi \text{ m}^2$ as calculated earlier. The reference to 'using integration' in the problem statement implies calculating the area of the inner ellipse using integration $4 \int_0^a y \, dx$ . In this case, $y = \frac{3}{4}\sqrt{16-x^2}$ . $4 \int_0^4 \frac{3}{4}\sqrt{16-x^2} \, dx = 3 \int_0^4 \sqrt{4^2-x^2} \, dx$ Using the formula $\int \sqrt{a^2 - x^2} \, dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right)$ . $3 \left[\frac{x}{2}\sqrt{16 - x^2} + \frac{16}{2}\sin^{-1}\left(\frac{x}{4}\right)\right]_0^4$ $= 3 \left[\left(\frac{4}{2}\sqrt{16 - 4^2} + 8\sin^{-1}\left(\frac{4}{4}\right)\right) - \left(\frac{0}{2}\sqrt{16 - 0^2} + 8\sin^{-1}\left(\frac{0}{4}\right)\right)\right]$ $= 3 \left[\left(2\sqrt{0} + 8\sin^{-1}(1)\right) - \left(0 + 8\sin^{-1}(0)\right)\right]$ $= 3 \left[8\left(\frac{\pi}{2}\right) - 0\right] = 3(4\pi) = 12\pi$

Step 9: Coordinates of P and Q for OR (iii) (b)

For part (iii) (b), we need the coordinates of points P and Q. Point P is where the outer ellipse cuts the $x$ -axis in the first quadrant, so its coordinates are ( $a_{outer}$ , 0). Point Q is where the outer ellipse cuts the $y$ -axis in the first quadrant, so its coordinates are (0, $b_{outer}$ ). The origin O is (0,0).

Step 10: Area of the triangle POQ using integration for OR (iii) (b)

To find the area of the triangle POQ using integration, we first need to define the line segment PQ. We find the equation of the line passing through P(5.5, 0) and Q(0, 4.5). Then, we integrate this linear function from $x=0$ to $x=5.5$ . The integral represents the area under the line segment PQ from the $x$ -axis to the line, which forms the triangle POQ. This confirms the result from the geometric area formula for a right-angled triangle.

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