We are given the equation of the ellipse as 9x² + 16y² = 144. To express y as a function of x, we need to isolate y on one side of the equation.

This integral is a standard form. We apply the formula for ∫√(a²-x²) dx, where a² = 16, so a=4. We will substitute this into our integral expression.

We substitute a=4 into the formula and multiply the entire expression by the constant factor (3)/(4). The constant of integration C is added since this is an indefinite integral.

First, we convert the given equation 9x² + 16y² = 144 into the standard form of an ellipse (x²)/(a²) + (y²)/(b²) = 1 by dividing by 144. This gives us (x²)/(16) + (y²)/(9) = 1. From this, we identify a² = 16 Rightarrow a = 4 and b² = 9 Rightarrow b = 3. The area of an ellipse is given by the formula π ab.

First, we convert the given equation 9x² + 16y² = 144 into the standard form of an ellipse (x²)/(a²) + (y²)/(b²) = 1 by dividing by 144. This gives us (x²)/(16) + (y²)/(9) = 1. From this, we identify a² = 16 Rightarrow a = 4 and b² = 9 Rightarrow b = 3. The area of an ellipse is given by the formula π ab.

The problem states that the track is around the elliptical ground. The image shows that the width of the track is added outside the existing ellipse. Therefore, the ellipse bounding the 'ground' part is the one given by the equation 9x² + 16y² = 144. The question asks for the area of the region 'enclosed within the elliptical ground excluding the track'. This means we need the area of the inner ellipse itself, which is what we just calculated.

The problem states that the track is around the elliptical ground. The image shows that the width of the track is added outside the existing ellipse. Therefore, the ellipse bounding the 'ground' part is the one given by the equation 9x² + 16y² = 144. The question asks for the area of the region 'enclosed within the elliptical ground excluding the track'. This means we need the area of the inner ellipse itself, which is what we just calculated.

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. (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) $12\pi \text{ m}^2$ (iii) (b) $P=(7, 0), Q=(0, 6)$ Area of $\triangle POQ = 21$ square units

Step-by-step solution

Step 1: Express y as a function of x

We are given the equation of the ellipse as $9x^2 + 16y^2 = 144$ . To express $y$ as a function of $x$ , we need to isolate $y$ on one side of the equation.

Step 2: Rearrange and solve for y

First, we subtract $9x^2$ from both sides. Then, we divide by 16 to get $y^2$ alone. Finally, we take the square root of both sides to find $y$ . We factor out $\sqrt{9/16} = 3/4$ from the expression inside the square root.

Step 3: Integrate the function

Now we need to integrate the function of $y$ with respect to $x$ . We use the positive root for the top half of the ellipse, as we are typically referring to the positive value of $y$ for area calculations.

Step 4: Apply standard integral formula

This integral is a standard form. We apply the formula for $\int\sqrt{a^2-x^2} \text{ d}x$ , where $a^2 = 16$ , so $a=4$ . We will substitute this into our integral expression.

Step 5: Substitute and simplify

We substitute $a=4$ into the formula and multiply the entire expression by the constant factor $\frac{3}{4}$ . The constant of integration $C$ is added since this is an indefinite integral.

Step 6: Find the area of the inner ellipse

First, we convert the given equation $9x^2 + 16y^2 = 144$ into the standard form of an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ by dividing by 144. This gives us $\frac{x^2}{16} + \frac{y^2}{9} = 1$ . From this, we identify $a^2 = 16 \Rightarrow a = 4$ and $b^2 = 9 \Rightarrow b = 3$ . The area of an ellipse is given by the formula $\pi ab$ .

Step 7: Determine the dimensions of the outer ellipse (track excluded)

The problem states that the track is around the elliptical ground. The image shows that the width of the track is added outside the existing ellipse. Therefore, the ellipse bounding the 'ground' part is the one given by the equation $9x^2 + 16y^2 = 144$ . The question asks for the area of the region 'enclosed within the elliptical ground excluding the track'. This means we need the area of the inner ellipse itself, which is what we just calculated.

Step 8: Calculate the area of the inner ellipse (final answer for part iii a)

The area calculated in the previous step, $12\pi \text{ m}^2$ , is the area of the elliptical ground. Since the question asks for the area of the region enclosed within the elliptical ground *excluding the track*, and the track is *around* the ground, this means we are simply looking for the area of the original ellipse. So, the area of the elliptical ground is the required area.

Step 9: Find coordinates of P and Q for OR part (iii) (b)

We first consider the original elliptical ground with equation $\frac{x^2}{16} + \frac{y^2}{9} = 1$ . The semi-major axis is $a=4$ and the semi-minor axis is $b=3$ . These values correspond to the intercepts of the original ellipse with the $x$ and $y$ axes.

Step 10: Determine dimensions of the outer ellipse

The track width is 3 meters. Since the track is around the elliptical ground, the dimensions of the outer ellipse (which represents the outer edge of the track) will be the dimensions of the inner ellipse plus the track width. The semi-major axis $a$ for the inner ellipse is 4 m and the semi-minor axis $b$ is 3 m. Adding the 3 m track width to both gives the new semi-major axis $a_{new} = 4+3 = 7$ m and new semi-minor axis $b_{new} = 3+3 = 6$ m.

Step 11: Determine coordinates of P and Q

Point P is where the outer edge of the track cuts the $x$ -axis in the first quadrant. This means P is $(a_{new}, 0)$ . Point Q is where the outer edge of the track cuts the $y$ -axis in the first quadrant, so Q is $(0, b_{new})$ . Using the new semi-axes $a_{new}=7$ and $b_{new}=6$ , we find the coordinates of P and Q.

Step 12: Calculate the area of triangle POQ

The triangle POQ has its vertices at the origin O(0,0), P(7,0) on the $x$ -axis, and Q(0,6) on the $y$ -axis. This is a right-angled triangle. The length of the base OP is 7 units and the height OQ is 6 units. The area of a right-angled triangle is half the product of its base and height.

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