Question

The area of the region bounded by \(\frac{x^{2}}{16} + \frac{y^{2}}{25} = 1\) and the line segment joining (0,5) and (4,0) in the first quadrant is

• A. \(10\pi -5\)
• B. \(5\pi -8\)
• C. \(4\pi -10\)
• D. \(4\pi -8\)
• E. \(5\pi -10\)

Hint:

Area of quarter ellipse = \(\frac{\pi ab}{4}\). Here \(a=4, b=5\), so quarter ellipse area = \(5\pi\).

Answer 1

Answer: (E)

Step 1: Concept:
• Given ellipse: \[ \frac{x^2}{16} + \frac{y^2}{25} = 1 \]
• In the first quadrant, required area = (area under ellipse) $-$ (area under the line).

Step 2: Detailed Explanation:
• Equation of ellipse: \[ y = 5\sqrt{1 - \frac{x^2}{16}} = \frac{5}{4}\sqrt{16 - x^2} \]
• Equation of line through \((0,5)\) and \((4,0)\): \[ y = -\frac{5}{4}x + 5 \]
• Required area: \[ \int_{0}^{4} \left[\frac{5}{4}\sqrt{16 - x^2} - \left(-\frac{5}{4}x + 5\right)\right] dx \]
• Split the integral: \[ = \frac{5}{4}\int_{0}^{4} \sqrt{16 - x^2} \, dx + \frac{5}{4}\int_{0}^{4} x \, dx - \int_{0}^{4} 5 \, dx \]
• Evaluate integrals: \[ \int_{0}^{4} \sqrt{16 - x^2} \, dx = \frac{1}{4}\pi(4^2) = 4\pi \] \[ \int_{0}^{4} x \, dx = 8 \] \[ \int_{0}^{4} 5 \, dx = 20 \]
• Substitute values: \[ \text{Area} = \frac{5}{4}(4\pi) + \frac{5}{4}(8) - 20 \]
• Simplify: \[ = 5\pi + 10 - 20 = 5\pi - 10 \]

Step 3: Final Answer:
• \[ \text{Area} = 5\pi - 10 \]