The ellipse `E_(1):(x^(2))/(9)+(y^(2))/(4)=1` is inscribed in a rectangle R whose sides are parallel to the coordinates axes. Another ellipse `E_(2)` passing through the point (0, 4) circumscribes the rectangle R. The length (in units) of the major axis of ellipse `E_(2)` is

Answer

To solve the problem, we need to analyze the two ellipses given: \( E_1 \) and \( E_2 \).

### Step 1: Identify the dimensions of the rectangle \( R \) inscribed by ellipse \( E_1 \).

The equation of the ellipse \( E_1 \) is given by:

\[ \frac{x^2}{9} + \frac{y^2}{4} = 1 \]

From this equation, we can identify the semi-major axis \( a \) and the semi-minor axis \( b \):

- \( a^2 = 9 \) → \( a = 3 \)
- \( b^2 = 4 \) → \( b = 2 \)

Thus, the rectangle \( R \) that inscribes the ellipse \( E_1 \) has dimensions:

- Width = \( 2a = 2 \times 3 = 6 \)
- Height = \( 2b = 2 \times 2 = 4 \)

### Step 2: Determine the coordinates of the vertices of rectangle \( R \).

The vertices of the rectangle \( R \) are at the points:

- \( (3, 2) \)
- \( (3, -2) \)
- \( (-3, 2) \)
- \( (-3, -2) \)

### Step 3: Analyze ellipse \( E_2 \).

Ellipse \( E_2 \) circumscribes rectangle \( R \) and passes through the point \( (0, 4) \). The general form of an ellipse that circumscribes a rectangle with width \( 2a \) and height \( 2b \) is:

\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]

Given that the rectangle has dimensions \( 6 \) (width) and \( 4 \) (height), we can say:

- \( a = 3 \)
- \( b = 2 \)

### Step 4: Find the relationship between the axes of ellipse \( E_2 \).

Since \( E_2 \) passes through the point \( (0, 4) \), we can substitute this point into the equation of ellipse \( E_2 \):

\[ \frac{0^2}{a^2} + \frac{4^2}{b^2} = 1 \]

This simplifies to:

\[ \frac{16}{b^2} = 1 \implies b^2 = 16 \implies b = 4 \]

### Step 5: Determine the semi-major axis \( a \) of ellipse \( E_2 \).

Since the rectangle is circumscribed by the ellipse, we can use the relationship derived from the rectangle's dimensions. The semi-major axis \( a \) must satisfy the condition:

\[ \frac{9}{a^2} + \frac{4}{16} = 1 \]

Substituting \( b^2 = 16 \):

\[ \frac{9}{a^2} + \frac{1}{4} = 1 \]

Rearranging gives:

\[ \frac{9}{a^2} = 1 - \frac{1}{4} = \frac{3}{4} \]

Cross-multiplying yields:

\[ 9 \cdot 4 = 3a^2 \implies 36 = 3a^2 \implies a^2 = 12 \implies a = \sqrt{12} = 2\sqrt{3} \]

### Step 6: Calculate the length of the major axis of ellipse \( E_2 \).

The length of the major axis of ellipse \( E_2 \) is given by:

\[ \text{Length of major axis} = 2a = 2 \times 2\sqrt{3} = 4\sqrt{3} \]

### Final Answer:

The length of the major axis of ellipse \( E_2 \) is \( 4\sqrt{3} \) units.

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