Question

If \( A_1, A_2, A_3 \) are the areas of the ellipse \( x^2 + 4y^2 = 4 \), its director circle, and auxiliary circle respectively, then \( A_2 + A_3 - A_1 \) is:

• A. \( 11\pi \)
• B. \( 3\pi \)
• C. \( 7\pi \)
• D. \( 9\pi \)

Hint:

For an ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), remember: - The area of the ellipse is \( \pi a b \). - The auxiliary circle has area \( \pi a^2 \). - The director circle has area \( \pi (a^2 + b^2) \).

Answer 1

The Correct Option is C

We are given the ellipse equation: \[ \frac{x^2}{4} + \frac{y^2}{1} = 1. \] Comparing with the standard ellipse equation \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), we identify: \[ a^2 = 4 \Rightarrow a = 2, \quad b^2 = 1 \Rightarrow b = 1. \] Step 1: Finding the area of the ellipse \( A_1 \)
The area of an ellipse is given by: \[ A_1 = \pi a b = \pi (2)(1) = 2\pi. \] Step 2: Finding the area of the auxiliary circle \( A_3 \)
The auxiliary circle has radius equal to the semi-major axis \( a = 2 \), so its area is: \[ A_3 = \pi a^2 = \pi (2)^2 = 4\pi. \] Step 3: Finding the area of the director circle \( A_2 \)
The director circle of an ellipse is given by radius \( \sqrt{a^2 + b^2} \), which simplifies to: \[ R = \sqrt{4 + 1} = \sqrt{5}. \] So, the area of the director circle is: \[ A_2 = \pi (\sqrt{5})^2 = 5\pi. \] Step 4: Calculating \( A_2 + A_3 - A_1 \)
\[ A_2 + A_3 - A_1 = 5\pi + 4\pi - 2\pi = 7\pi. \]