Question

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

Answer 1

Concept:
• Transposing terms to isolate one variable in terms of the other.

Step 1: Isolate the term containing \( y \)
Given equation: \( 9x^2 + 16y^2 = 144 \)
\[ 16y^2 = 144 - 9x^2 \]

Step 2: Solve for \( y \)
Divide by 16:
\[ y^2 = \frac{144 - 9x^2}{16} = 9 - \frac{9}{16}x^2 \]
\[ y^2 = \frac{9}{16}(16 - x^2) \]
Taking square root (considering the positive branch for a function):
\[ y = \frac{3}{4} \sqrt{16 - x^2} \]

Answer 2

Concept:
• Standard Integral: \( \int \sqrt{a^2 - x^2} dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right) + C \).

Step 1: Set up the integral
\[ I = \int \frac{3}{4} \sqrt{16 - x^2} dx \]
Here \( a^2 = 16 \), so \( a = 4 \).

Step 2: Apply the standard formula
\[ I = \frac{3}{4} \left[ \frac{x}{2}\sqrt{16 - x^2} + \frac{16}{2}\sin^{-1}\left(\frac{x}{4}\right) \right] + C \]
\[ I = \frac{3}{4} \left[ \frac{x}{2}\sqrt{16 - x^2} + 8\sin^{-1}\left(\frac{x}{4}\right) \right] + C \]

Step 3: Simplify the expression
\[ I = \frac{3}{8}x\sqrt{16 - x^2} + 6\sin^{-1}\left(\frac{x}{4}\right) + C \]

Answer 3

Concept:

• Area of ellipse is 4 times the area in the first quadrant.
• Area \( = 4 \int_{0}^{a} y dx \).

Step 1: Determine the limits
The standard form is \( \frac{x^2}{16} + \frac{y^2}{9} = 1 \), so \( a=4 \).
The limits for the first quadrant are from 0 to 4.

Step 2: Calculate the definite integral
Using the result from (ii):
\[ \text{Area} = 4 \int_{0}^{4} \frac{3}{4} \sqrt{16 - x^2} dx \]
\[ \text{Area} = 4 \left[ \frac{3}{8}x\sqrt{16 - x^2} + 6\sin^{-1}\left(\frac{x}{4}\right) \right]_{0}^{4} \]

Step 3: Substitute values
Upper limit (4): \( \frac{3}{8}(4)\sqrt{0} + 6\sin^{-1}(1) = 0 + 6(\pi/2) = 3\pi \)
Lower limit (0): \( 0 + 6\sin^{-1}(0) = 0 \)
Total Area \( = 4 \times 3\pi = 12\pi \text{ sq units} \).

Answer 4

Concept:
• Expanding the semi-axes by adding the track width.
• Equation of a line in intercept form: \( \frac{x}{a} + \frac{y}{b} = 1 \).

Step 1: Find coordinates of P and Q
Inner semi-axes: \( a=4, b=3 \).
Width = 3 m. Thus outer semi-axes are \( a' = 4+3 = 7 \) and \( b' = 3+3 = 6 \).
Outer edge cuts x-axis at \( P(7, 0) \) and y-axis at \( Q(0, 6) \).

Step 2: Find the equation of line PQ
Using intercept form \( \frac{x}{7} + \frac{y}{6} = 1 \):
\[ y = 6 \left( 1 - \frac{x}{7} \right) = 6 - \frac{6x}{7} \]

Step 3: Find the area using integration
\[ \text{Area} = \int_{0}^{7} \left( 6 - \frac{6x}{7} \right) dx \]
\[ \text{Area} = \left[ 6x - \frac{3x^2}{7} \right]_{0}^{7} \]
\[ \text{Area} = (42 - 21) - (0) = 21 \text{ sq units} \]