Divide both sides of the equation 9x² + 16y² = 144 by 144. This simplifies the equation to the standard form of an ellipse, (x²)/(a²) + (y²)/(b²) = 1. From this, we can identify a² = 16 and b² = 9 for the inner ellipse.

Divide both sides of the equation 9x² + 16y² = 144 by 144. This simplifies the equation to the standard form of an ellipse, (x²)/(a²) + (y²)/(b²) = 1. From this, we can identify a² = 16 and b² = 9 for the inner ellipse.

Now, we isolate y² on one side of the equation. Then, we take the square root of both sides to express y as a function of x. Since y can be positive or negative, we include the ± sign. We simplify the expression under the square root to get the final form of y in terms of x for the upper and lower halves of the ellipse.

Now, we isolate y² on one side of the equation. Then, we take the square root of both sides to express y as a function of x. Since y can be positive or negative, we include the ± sign. We simplify the expression under the square root to get the final form of y in terms of x for the upper and lower halves of the ellipse.

We are asked to integrate the function obtained in part (i) with respect to x. We use the positive root y = (3)/(4) √(16 - x²) for integration. This integral is of the form ∫ √(a² - x²) dx, which is a standard integral formula that will be applied in the next steps.

We are asked to integrate the function obtained in part (i) with respect to x. We use the positive root y = (3)/(4) √(16 - x²) for integration. This integral is of the form ∫ √(a² - x²) dx, which is a standard integral formula that will be applied in the next steps.

Substitute a=4 into the formula and multiply by the constant (3)/(4). This gives the indefinite integral of the function y = (3)/(4) √(16 - x²).

Now we evaluate the definite integral by plugging in the upper limit (x=4) and the lower limit (x=0). We know that sin^(-1)(1) = (π)/(2) and sin^(-1)(0) = 0. This gives us the total area of the inner elliptical ground.

Now we evaluate the definite integral by plugging in the upper limit (x=4) and the lower limit (x=0). We know that sin^(-1)(1) = (π)/(2) and sin^(-1)(0) = 0. This gives us the total area of the inner elliptical ground.

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, racing track)

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) $12\pi \text{ units}^2$ (iii) (b) Coordinates P $(7{,}0)$ , Q $(0{,}6)$ , Area $21 \text{ units}^2$

Step-by-step solution

Step 1: Standard form of Ellipse

To express $y$ as a function of $x$ , we first convert the given equation of the ellipse into its standard form. This involves dividing the entire equation by the constant term on the right side to make it equal to 1.

Step 2: Simplify Ellipse Equation

Divide both sides of the equation $9x^2 + 16y^2 = 144$ by 144. This simplifies the equation to the standard form of an ellipse, $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ . From this, we can identify $a^2 = 16$ and $b^2 = 9$ for the inner ellipse.

Step 3: Express y in terms of x

Now, we isolate $y^2$ on one side of the equation. Then, we take the square root of both sides to express $y$ as a function of $x$ . Since $y$ can be positive or negative, we include the $\pm$ sign. We simplify the expression under the square root to get the final form of $y$ in terms of $x$ for the upper and lower halves of the ellipse.

Step 4: Integrate the Function (i)

We are asked to integrate the function obtained in part (i) with respect to $x$ . We use the positive root $y = \frac{3}{4} \sqrt{16 - x^2}$ for integration. This integral is of the form $\int \sqrt{a^2 - x^2} \, dx$ , which is a standard integral formula that will be applied in the next steps.

Step 5: Apply Standard Integral Formula

Recall the standard formula for integrating $\sqrt{a^2 - x^2}$ . Here, $a=4$ . We will substitute this into the formula. This formula is commonly used when finding the area under curves defined by elliptical or circular segments.

Step 6: Substitute into the integral

Substitute $a=4$ into the formula and multiply by the constant $\frac{3}{4}$ . This gives the indefinite integral of the function $y = \frac{3}{4} \sqrt{16 - x^2}$ .

Step 7: Find Area of Inner Ellipse (Part iii a)

The total area of an ellipse is symmetric across both axes. We can calculate the area of the part in the first quadrant by integrating $y$ from $0$ to $a$ (which is $4$ ) and then multiply the result by $4$ . We use the integral form obtained in the previous step and apply the limits of integration from $0$ to $4$ .

Step 8: Evaluate definite integral for inner ellipse

Now we evaluate the definite integral by plugging in the upper limit ( $x=4$ ) and the lower limit ( $x=0$ ). We know that $\sin^{-1}(1) = \frac{\pi}{2}$ and $\sin^{-1}(0) = 0$ . This gives us the total area of the inner elliptical ground.

Step 9: Calculate outer ellipse dimensions

The track has a uniform width of 3 m. This means the semi-major axis and semi-minor axis of the outer ellipse will be 3 m larger than those of the inner ellipse. For the inner ellipse, $a = \sqrt{16} = 4$ and $b = \sqrt{9} = 3$ . So, for the outer ellipse, the new semi-major axis $a_o$ will be $4+3=7$ and the new semi-minor axis $b_o$ will be $3+3=6$ .

Step 10: Find Area of Outer Ellipse

The formula for the area of an ellipse is $\pi ab$ . Using the new dimensions for the outer ellipse ( $a_o=7$ and $b_o=6$ ), we can calculate its total area.

Step 11: Area of Region Excluding Track (Part iii a)

The problem asks for 'the area of the region enclosed within the elliptical ground excluding the track'. This refers to the area of the inner ellipse, which we have already calculated. The 'track' is the region *between* the inner and outer ellipses.

Step 12: Coordinates of P and Q (Part iii b)

For part (iii) (b), we need the coordinates of points P and Q on the outer edge of the track, in the first quadrant. P is the x-intercept, so $y=0$ . Q is the y-intercept, so $x=0$ . Using the semi-axes of the outer ellipse, $a_o=7$ and $b_o=6$ , point P is $(7{,}0)$ and point Q is $(0{,}6)$ . Point O is the origin $(0{,}0)$ .

Step 13: Area of Triangle POQ using Integration

To find the area of the triangle POQ using integration, we need the equation of the line segment PQ. The integral will be evaluated from $x=0$ to $x=7$ (the x-coordinate of P) of the function representing line PQ.

Step 14: Equation of Line PQ

Using the intercept form for a line, $\frac{x}{a} + \frac{y}{b} = 1$ , where $a$ is the x-intercept and $b$ is the y-intercept. For line PQ, the x-intercept is $7$ and the y-intercept is $6$ . We rearrange this equation to express $y$ as a function of $x$ .

Step 15: Integrate to find Triangle Area

Finally, we evaluate the definite integral of the line equation from $x=0$ to $x=7$ . This calculates the area under the line segment PQ, which is the area of the triangle POQ. This confirms the standard formula for the area of a right-angled triangle: $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 7 \times 6 = 21$ .

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