Question:medium

Area lying in the first quadrant and bounded by ellipse \(4x^2 + 9y^2 = 144\) is _____

Show Hint

Area in first quadrant of symmetric curves like ellipse = one-fourth of total area.
Updated On: Apr 2, 2026
  • \( 24\pi \)
  • \( 8\pi \)
  • \( 12\pi \)
  • \( 6\pi \)
Show Solution

The Correct Option is D

Solution and Explanation

Step 1: Understanding the Concept:
The standard equation of an ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. The total area is $\pi ab$. Since we only need the first quadrant, we take $1/4$ of the total area.
Step 2: Formula Application:
Divide the given equation $4x^2 + 9y^2 = 144$ by 144: $$\frac{4x^2}{144} + \frac{9y^2}{144} = 1 \implies \frac{x^2}{36} + \frac{y^2}{16} = 1$$ Here, $a^2 = 36 \implies a = 6$ and $b^2 = 16 \implies b = 4$.
Step 3: Explanation:
Total Area = $\pi \times 6 \times 4 = 24\pi$. Area in the first quadrant = $\frac{1}{4} \times 24\pi = 6\pi$.
Step 4: Final Answer:
The area is $6\pi$.
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