If the line joining foci subtends an angle of `90^(@)` at an extremity of minor axis, then the eccentricity e is

Options

A. `(1)/(sqrt(6))`
B. `(1)/(sqrt(3))`
C. `(1)/(sqrt(2))`
D. `(1)/(2sqrt(2))`

Answer

To solve the problem, we need to find the eccentricity \( e \) of the ellipse given that the line joining the foci subtends a \( 90^\circ \) angle at an extremity of the minor axis. Here’s the step-by-step solution:

### Step 1: Understand the Geometry of the Ellipse
An ellipse has two foci, denoted as \( F_1 \) and \( F_2 \). The major axis is the longest diameter of the ellipse, and the minor axis is the shortest. The foci lie along the major axis, and the endpoints of the minor axis are perpendicular to the major axis.

**Hint:** Visualize the ellipse and label the foci and axes correctly.

### Step 2: Define the Points and Angles
Let \( P \) be the extremity of the minor axis. According to the problem, the angle \( F_1PF_2 \) is \( 90^\circ \). This means that the triangle \( F_1PF_2 \) is a right triangle.

**Hint:** Remember that in a right triangle, the Pythagorean theorem applies.

### Step 3: Use the Definition of the Ellipse
By the definition of an ellipse, the sum of the distances from any point on the ellipse to the two foci is constant and equal to \( 2a \), where \( a \) is the semi-major axis. Therefore, we have:
\[ PF_1 + PF_2 = 2a \]

**Hint:** Identify \( PF_1 \) and \( PF_2 \) in terms of \( a \).

### Step 4: Apply the Pythagorean Theorem
Since \( F_1PF_2 \) is a right triangle, we can apply the Pythagorean theorem:
\[ PF_1^2 + PF_2^2 = F_1F_2^2 \]
Let \( PF_1 = a \) and \( PF_2 = a \) (since both distances are equal at the extremity of the minor axis). The distance between the foci \( F_1F_2 \) is \( 2ae \), where \( e \) is the eccentricity.

Thus, we can write:
\[ a^2 + a^2 = (2ae)^2 \]
This simplifies to:
\[ 2a^2 = 4a^2e^2 \]

**Hint:** Simplify the equation carefully.

### Step 5: Solve for Eccentricity \( e \)
Dividing both sides by \( 2a^2 \) (assuming \( a \neq 0 \)):
\[ 1 = 2e^2 \]
Thus, we can solve for \( e^2 \):
\[ e^2 = \frac{1}{2} \]
Taking the square root gives:
\[ e = \frac{1}{\sqrt{2}} \]

**Hint:** Remember that eccentricity \( e \) must be a positive value.

### Final Answer
The eccentricity \( e \) is:
\[ e = \frac{1}{\sqrt{2}} \]