If the circle whose diameter is the major axis of the ellipse `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1(a gt b gt 0)` meets the minor axis at point P and the orthocentre of `DeltaPF_(1)F_(2)` lies on the ellipse, where `F_(1) and F_(2)` are foci of the ellipse, then the square of the eccentricity of the ellipse is

Options

A. `(sqrt5-1)/(2)`
B. `sqrt3-1`
C. `(1)/(sqrt2)`
D. `(sqrt3)/(2)`

Answer

To solve the problem, we need to analyze the given ellipse and its properties. The ellipse is defined by the equation:

\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \quad (a > b > 0) \]

The foci of the ellipse, \(F_1\) and \(F_2\), are located at \((\pm c, 0)\), where \(c = \sqrt{a^2 - b^2}\).

### Step 1: Identify the Circle
The circle whose diameter is the major axis of the ellipse has its center at the origin and its radius equal to half the length of the major axis, which is \(a\). Therefore, the equation of the circle is:

\[ x^2 + y^2 = a^2 \]

### Step 2: Find Intersection Points with the Minor Axis
The minor axis of the ellipse is the line \(x = 0\). To find the intersection points of the circle with the minor axis, we substitute \(x = 0\) into the circle's equation:

\[ 0^2 + y^2 = a^2 \implies y^2 = a^2 \implies y = \pm a \]

Thus, the points of intersection are \(P(0, a)\) and \(P(0, -a)\).

### Step 3: Determine the Orthocenter of Triangle \(PF_1F_2\)
The vertices of triangle \(PF_1F_2\) are:
- \(P(0, a)\)
- \(F_1(-c, 0)\)
- \(F_2(c, 0)\)

To find the orthocenter of triangle \(PF_1F_2\), we need to find the altitudes of the triangle.

### Step 4: Calculate the Slopes of the Sides
1. Slope of \(PF_1\):
\[ m_{PF_1} = \frac{0 - a}{-c - 0} = \frac{-a}{-c} = \frac{a}{c} \]

2. Slope of \(PF_2\):
\[ m_{PF_2} = \frac{0 - a}{c - 0} = \frac{-a}{c} \]

3. Slope of \(F_1F_2\):
\[ m_{F_1F_2} = \frac{0 - 0}{c - (-c)} = 0 \]

### Step 5: Find the Altitudes
The altitude from \(P\) to \(F_1F_2\) is a vertical line at \(x = 0\).

The altitude from \(F_1\) to \(PF_2\) has a slope of \(-\frac{c}{a}\) (negative reciprocal of \(PF_2\)). The equation of the line through \(F_1(-c, 0)\) is:

\[ y - 0 = -\frac{c}{a}(x + c) \implies y = -\frac{c}{a}x - \frac{c^2}{a} \]

### Step 6: Find the Orthocenter
The orthocenter lies at the intersection of the altitudes. We can set the equations equal to find the coordinates of the orthocenter.

### Step 7: Condition for Orthocenter to Lie on the Ellipse
For the orthocenter to lie on the ellipse, we substitute its coordinates into the ellipse equation and simplify to find a relation involving \(a\) and \(b\).

### Step 8: Eccentricity Calculation
The eccentricity \(e\) of the ellipse is given by:

\[ e = \frac{c}{a} = \frac{\sqrt{a^2 - b^2}}{a} \]

The square of the eccentricity is:

\[ e^2 = \frac{a^2 - b^2}{a^2} \]

### Final Result
Thus, the square of the eccentricity of the ellipse is:

\[ \boxed{\frac{a^2 - b^2}{a^2}} \]