CUET · MATHS · PYQ PAPER 2025
The area of the smaller region bounded by the ellipse \(\frac{x^2}{16}+\frac{y^2}{9}=1\) and the straight line \(3 x+4 y=12\) is :
- A \((6 \pi-3)\) sq units
- B \((3 \pi+6)\) sq units
- C \((6 \pi+3)\) sq units
- D \((3 \pi-6)\) sq units
Answer & Solution
Correct Answer
(D) \((3 \pi-6)\) sq units
Step-by-step Solution
Detailed explanation
Equation of ellipse: \(\frac{x^2}{4^2}+\frac{y^2}{3^2}=1\). Line \(3x+4y=12\) intersects the ellipse at \((4,0)\) and \((0,3)\). Area of ellipse in the first quadrant = \(\frac{1}{4} \pi a b = \frac{1}{4} \pi (4)(3) = 3\pi\). Area of triangle formed by the line and coordinate…
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