Question

Express \(y\) as a function of \(x\) from the given equation of ellipse.

Answer 1

Concept: An equation expresses one variable explicitly in terms of another when it is rearranged into the form \(y = f(x)\). To isolate \(y\) from a general quadratic relation of an ellipse such as \(Ax^2 + By^2 = C\), we follow standard algebraic operations:
• Subtract the term containing \(x\) from both sides of the equation.
• Divide the entire equation by the coefficient of \(y^2\) to isolate the squared term.
• Take the square root of both sides, remembering to include both the positive and negative roots (\(\pm\)) since squaring either a positive or a negative real number yields a positive result.

Step 1: Write down the given equation and isolate the \(y^2\) term.
The given equation representing the inner boundary of the elliptical ground is: \[ 9x^2 + 16y^2 = 144 \] To express \(y\) as a function of \(x\), we first need to isolate the term containing \(y\), which is \(16y^2\). We achieve this by subtracting \(9x^2\) from both sides of the equation: \[ 16y^2 = 144 - 9x^2 \]

Step 2: Factor out common terms and isolate \(y^2\).
On the right-hand side, we can notice that \(144\) is a multiple of \(9\) because \(9 \times 16 = 144\). Let us explicitly factor out \(9\) from the right-hand side expression to make subsequent simplification cleaner: \[ 16y^2 = 9(16 - x^2) \] Now, to fully isolate \(y^2\), we divide both sides of the equation by its coefficient, which is \(16\): \[ y^2 = \frac{9(16 - x^2)}{16} \] This can also be written split up as: \[ y^2 = \frac{9}{16}(16 - x^2) \]

Step 3: Solve for \(y\) by taking the square root.
To find \(y\), we take the square root of both sides of the equation. Because both a positive number and its negative counterpart yield the same value when squared, we must introduce the plus-minus sign (\(\pm\)): \[ y = \pm \sqrt{\frac{9}{16}(16 - x^2)} \] Using the algebraic property of radicals, \(\sqrt{A \cdot B} = \sqrt{A} \cdot \sqrt{B}\), we can separate the constant fractional part from the algebraic expression inside the square root: \[ y = \pm \sqrt{\frac{9}{16}} \cdot \sqrt{16 - x^2} \] Since \(9 = 3^2\) and \(16 = 4^2\), the square root of the fraction \(\frac{9}{16}\) is a perfect rational value: \[ \sqrt{\frac{9}{16}} = \frac{\sqrt{9}}{\sqrt{16}} = \frac{3}{4} \] Substituting this back into our equation gives the final explicit functional relation for \(y\): \[ y = \pm \frac{3}{4}\sqrt{16 - x^2} \] This perfectly matches option (A).

Answer 2

Concept: To find the indefinite integral of an algebraic function involving a square root of the form \(\sqrt{a^2 - x^2}\), we employ the standard trigonometric substitution or the direct integration formula derived via integration by parts. The standard integration formula is: \[ \int \sqrt{a^2 - x^2} \, dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right) + C \] where \(a\) is a positive real constant and \(C\) is the arbitrary constant of integration.

Step 1: Set up the integral expression.
From the previous part (i), the explicit function obtained for \(y\) in terms of \(x\) is: \[ y = \pm \frac{3}{4}\sqrt{16 - x^2} \] We are required to compute the indefinite integral of this function with respect to \(x\), which is represented mathematically as: \[ \int y \, dx = \int \left( \pm \frac{3}{4}\sqrt{16 - x^2} \right) dx \] By the linearity property of integration, constants can be factored out completely in front of the integral sign: \[ \int y \, dx = \pm \frac{3}{4} \int \sqrt{16 - x^2} \, dx \]

Step 2: Apply the standard trigonometric integral formula.
Observe the term inside the square root: \(16 - x^2\). This is of the form \(a^2 - x^2\), where: \[ a^2 = 16 \quad \Rightarrow \quad a = 4 \] Now, substituting \(a = 4\) into the standard standard formula \(\int \sqrt{a^2 - x^2} \, dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right)\), we get: \[ \int \sqrt{16 - x^2} \, dx = \frac{x}{2}\sqrt{16 - x^2} + \frac{16}{2}\sin^{-1}\left(\frac{x}{4}\right) \] Simplifying the coefficient of the inverse sine term, since \(\frac{16}{2} = 8\): \[ \int \sqrt{16 - x^2} \, dx = \frac{x}{2}\sqrt{16 - x^2} + 8\sin^{-1}\left(\frac{x}{4}\right) \]

Step 3: Combine with the factored constant and add the constant of integration.
Now multiply this complete expression by the constant factor \(\pm \frac{3}{4}\) that was pulled outside the integral: \[ \int y \, dx = \pm \frac{3}{4} \left[ \frac{x}{2}\sqrt{16 - x^2} + 8\sin^{-1}\left(\frac{x}{4}\right) \right] + C \] where \(C\) is the constant of integration. This matches option (A) perfectly.

Answer 3

Concept: The inner boundary of the track represents the elliptical ground. The phrase "region enclosed within the elliptical ground" simply means computing the total internal area of this ellipse. Since an ellipse is perfectly symmetric about both the \(x\)-axis and the \(y\)-axis, its total area can be computed by taking four times the area located inside the first quadrant. The boundaries in the first quadrant run along the \(x\)-axis from \(x = 0\) to the positive \(x\)-intercept (semi-major axis \(a\)). The definite integral formulation is: \[ \text{Total Area} = 4 \times \int_{0}^{a} y_{\text{positive}} \, dx \]

Step 1: Convert the ellipse equation into standard form to find the limits.
The given equation of the inner elliptical ground is: \[ 9x^2 + 16y^2 = 144 \] To clearly locate the limits of integration, divide both sides by \(144\): \[ \frac{9x^2}{144} + \frac{16y^2}{144} = 1 \] Simplifying the fractions: \[ \frac{x^2}{16} + \frac{y^2}{9} = 1 \quad \Rightarrow \quad \frac{x^2}{4^2} + \frac{y^2}{3^2} = 1 \] Comparing this to the standard ellipse equation \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), we find:
• Semi-major axis \(a = 4\) (the curve cuts the positive \(x\)-axis at \(x = 4\))
• Semi-minor axis \(b = 3\) (the curve cuts the positive \(y\)-axis at \(y = 3\)) Hence, in the first quadrant, \(x\) ranges from \(0\) to \(4\).

Step 2: Set up the definite integral for the area.
In the first quadrant, \(y\) is strictly positive, so we select the positive branch from part (i): \[ y = \frac{3}{4}\sqrt{16 - x^2} \] The total area \(A\) is given by four times the area under this curve from \(x=0\) to \(x=4\): \[ A = 4 \times \int_{0}^{4} \frac{3}{4}\sqrt{16 - x^2} \, dx \] We can cancel out the constant factor of \(4\) in the numerator and denominator: \[ A = 3 \int_{0}^{4} \sqrt{16 - x^2} \, dx \]

Step 3: Evaluate the definite integral using limits.
Utilizing the integration result from part (ii): \[ A = 3 \left[ \frac{x}{2}\sqrt{16 - x^2} + \frac{16}{2}\sin^{-1}\left(\frac{x}{4}\right) \right]_{0}^{4} \] \[ A = 3 \left[ \frac{x}{2}\sqrt{16 - x^2} + 8\sin^{-1}\left(\frac{x}{4}\right) \right]_{0}^{4} \] Now we carefully substitute the upper limit (\(x = 4\)) and lower limit (\(x = 0\)):
• At upper limit \(x = 4\): \[ \text{Value} = \frac{4}{2}\sqrt{16 - 4^2} + 8\sin^{-1}\left(\frac{4}{4}\right) = 2\sqrt{16 - 16} + 8\sin^{-1}(1) \] Since \(\sqrt{0} = 0\) and \(\sin^{-1}(1) = \frac{\pi}{2}\): \[ \text{Value} = 2(0) + 8\left(\frac{\pi}{2}\right) = 4\pi \]
• At lower limit \(x = 0\): \[ \text{Value} = \frac{0}{2}\sqrt{16 - 0^2} + 8\sin^{-1}\left(\frac{0}{4}\right) = 0 + 8\sin^{-1}(0) \] Since \(\sin^{-1}(0) = 0\): \[ \text{Value} = 0 \] Subtracting the lower limit value from the upper limit value: \[ A = 3 \cdot (4\pi - 0) = 12\pi \text{ m}^2 \] The calculated area matches option (B).

Answer 4

Concept: The track is built uniformly around the inner elliptical ground. The width of this track is given as \(3\text{ m}\).
• The inner boundary intersects the positive \(x\)-axis at a distance equal to its semi-major axis \(a = 4\). Therefore, the outer boundary point \(\text{P}\) on the positive \(x\)-axis is shifted outwards by adding the width of the track.
• Similarly, the inner boundary intersects the positive \(y\)-axis at a distance equal to its semi-minor axis \(b = 3\). The outer boundary point \(\text{Q}\) on the positive \(y\)-axis is shifted upwards by adding the width of the track. Once coordinates \(\text{P}(x_1, 0)\) and \(\text{Q}(0, y_1)\) are established, the equation of line \(\text{PQ}\) can be written. The area of \(\triangle\text{POQ}\) is then computed via integration by integrating the equation of line \(\text{PQ}\) from \(x = 0\) to \(x = x_1\).

Step 1: Determine the coordinates of points P and Q.
From our previous analysis of the inner ellipse, the inner intercepts are: \[ \text{Inner } x\text{-intercept} = 4, \quad \text{Inner } y\text{-intercept} = 3 \] Given that the track has a uniform width of \(3\text{ m}\) surrounding the inner ground: \[ \text{Outer } x\text{-intercept for point P} = \text{Inner } x\text{-intercept} + \text{width} = 4 + 3 = 7 \] Since point \(\text{P}\) lies on the positive \(x\)-axis, its \(y\)-coordinate is strictly \(0\). Thus: \[ \text{P} = (7, 0) \] Similarly, for point \(\text{Q}\) lying on the positive \(y\)-axis: \[ \text{Outer } y\text{-intercept for point Q} = \text{Inner } y\text{-intercept} + \text{width} = 3 + 3 = 6 \] Since point \(\text{Q}\) lies on the positive \(y\)-axis, its \(x\)-coordinate is strictly \(0\). Thus: \[ \text{Q} = (0, 6) \]

Step 2: Find the equation of the line segment PQ.
The line passes through \(\text{Q}(0,6)\) and \(\text{P}(7,0)\). We can write the equation of the line using the intercept form \(\frac{x}{x_{\text{int}}} + \frac{y}{y_{\text{int}}} = 1\): \[ \frac{x}{7} + \frac{y}{6} = 1 \] Now, let us rearrange this equation to isolate \(y\) as a function of \(x\): \[ \frac{y}{6} = 1 - \frac{x}{7} \] \[ \frac{y}{6} = \frac{7 - x}{7} \] Multiplying both sides by \(6\): \[ y = \frac{6}{7}(7 - x) \quad \Rightarrow \quad y = 6 - \frac{6}{7}x \]

Step 3: Integrate to find the area of triangle POQ.
The triangle \(\text{POQ}\) is bounded above by the line \(y = \frac{6}{7}(7-x)\), below by the \(x\)-axis (\(y=0\)), on the left by the \(y\)-axis (\(x=0\)), and on the right by the point \(\text{P}\) (\(x=7\)). The area \(A_{\triangle}\) is found by integrating from \(x = 0\) to \(x = 7\): \[ A_{\triangle} = \int_{0}^{7} \frac{6}{7}(7 - x) \, dx \] Factoring out the constant fraction \(\frac{6}{7}\): \[ A_{\triangle} = \frac{6}{7} \int_{0}^{7} (7 - x) \, dx \] Performing the integration term by term: \[ A_{\triangle} = \frac{6}{7} \left[ 7x - \frac{x^2}{2} \right]_{0}^{7} \] Now, substitute the upper limit (\(x = 7\)) and lower limit (\(x = 0\)):
• At upper limit \(x = 7\): \[ \text{Value} = 7(7) - \frac{7^2}{2} = 49 - \frac{49}{2} = \frac{98 - 49}{2} = \frac{49}{2} \]
• At lower limit \(x = 0\): \[ \text{Value} = 7(0) - \frac{0^2}{2} = 0 \] Subtracting the results: \[ A_{\triangle} = \frac{6}{7} \times \left( \frac{49}{2} - 0 \right) \] \[ A_{\triangle} = \frac{6 \times 49}{7 \times 2} \] Simplifying by dividing terms: \[ A_{\triangle} = \left(\frac{6}{2}\right) \times \left(\frac{49}{7}\right) = 3 \times 7 = 21 \text{ m}^2 \] Thus, the coordinates are \(\text{P}(7,0), \text{Q}(0,6)\) and the area of the triangle is \(21 \text{ m}^2\), matching option (A).