Consider the ellipse `E_1, x^2/a^2+y^2/b^2=1,(a>b)`. An ellipse `E_2` passes through the extremities of the major axis of `E_1` and has its foci at the ends of its minor axis.Consider the following property:Sum of focal distances of any point on an ellipse is equal to its major axis. Equation of `E_2` is

Options

A. `2:1`
B. `3:2`
C. `sqrt2:1`
D. `5:2`

Explanation

To solve the problem, we need to find the equation of the ellipse \( E_2 \) based on the properties of the ellipse \( E_1 \). Let's break down the steps.

### Step 1: Identify the properties of ellipse \( E_1 \)
The equation of the ellipse \( E_1 \) is given by:

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

where \( a > b \). The major axis is along the x-axis, and the foci are located at \( (\pm ae, 0) \) where \( e = \sqrt{1 - \frac{b^2}{a^2}} \).

### Step 2: Determine the extremities of the major axis of \( E_1 \)
The extremities of the major axis of \( E_1 \) are at the points \( (a, 0) \) and \( (-a, 0) \).

### Step 3: Identify the foci of ellipse \( E_2 \)
The foci of ellipse \( E_2 \) are located at the ends of the minor axis of \( E_1 \), which are at the points \( (0, b) \) and \( (0, -b) \).

### Step 4: Use the property of ellipse \( E_2 \)
According to the property of ellipses, the sum of the distances from any point on the ellipse to the two foci is equal to the length of the major axis. For ellipse \( E_2 \), the length of the major axis is \( 2a' \), where \( a' \) is the semi-major axis.

### Step 5: Set up the equation for ellipse \( E_2 \)
Let the semi-major axis of \( E_2 \) be \( a' \) and the semi-minor axis be \( b' \). The foci are at \( (0, b) \) and \( (0, -b) \), so the distance from any point \( (x, y) \) on the ellipse to the foci is given by:

\[ \sqrt{x^2 + (y - b)^2} + \sqrt{x^2 + (y + b)^2} = 2a' \]

### Step 6: Find the relationship between \( a' \) and \( b' \)
For ellipse \( E_2 \), the relationship between the semi-major axis \( a' \) and the semi-minor axis \( b' \) is given by:

\[ c^2 = a'^2 - b'^2 \]

where \( c \) is the distance from the center to each focus, which is equal to \( b \) (the semi-minor axis of \( E_1 \)). Thus, we have:

\[ b^2 = a'^2 - b'^2 \]

### Step 7: Solve for the equation of \( E_2 \)
The equation of ellipse \( E_2 \) can be expressed as:

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

### Step 8: Substitute \( a' \) and \( b' \)
From the previous steps, we can express \( a' \) and \( b' \) in terms of \( a \) and \( b \). We know that:

1. The semi-major axis \( a' = b \) (the length of the minor axis of \( E_1 \)).
2. The semi-minor axis \( b' = a \) (the length of the major axis of \( E_1 \)).

Thus, the equation of \( E_2 \) becomes:

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

### Final Answer
The equation of the ellipse \( E_2 \) is:

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