If `x^2/a^2+y^2/b^2=1` passes through `(sqrt(3/2),1) , e=1/sqrt3` , circle centred at one of the focus and radius `2/sqrt3`. these ellipse and circle intersect at two points . Find square of the distance between the two points is

Options

A. `4/3`
B. `2/3`
C. `16/3`
D. `32/3`

Answer

To solve the problem step by step, we will follow the given conditions and equations.

### Step 1: Understand the ellipse equation and its parameters
The equation of the ellipse is given by:

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

We know the eccentricity \( e \) of the ellipse is given by:

\[ e = \sqrt{1 - \frac{b^2}{a^2}} \]

Given \( e = \frac{1}{\sqrt{3}} \), we can square this to find:

\[ e^2 = \frac{1}{3} = 1 - \frac{b^2}{a^2} \]

Rearranging gives us:

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

Thus, we can express \( b^2 \) in terms of \( a^2 \):

\[ b^2 = \frac{2}{3} a^2 \]

### Step 2: Use the point through which the ellipse passes
The ellipse passes through the point \( \left(\sqrt{\frac{3}{2}}, 1\right) \). Substituting this point into the ellipse equation gives:

\[ \frac{\left(\sqrt{\frac{3}{2}}\right)^2}{a^2} + \frac{1^2}{b^2} = 1 \]

Calculating \( \left(\sqrt{\frac{3}{2}}\right)^2 = \frac{3}{2} \):

\[ \frac{\frac{3}{2}}{a^2} + \frac{1}{b^2} = 1 \]

Substituting \( b^2 = \frac{2}{3} a^2 \):

\[ \frac{\frac{3}{2}}{a^2} + \frac{1}{\frac{2}{3} a^2} = 1 \]

This simplifies to:

\[ \frac{\frac{3}{2}}{a^2} + \frac{3}{2a^2} = 1 \]

Combining the fractions:

\[ \frac{3 + 3}{2a^2} = 1 \implies \frac{6}{2a^2} = 1 \implies \frac{3}{a^2} = 1 \implies a^2 = 3 \]

### Step 3: Find \( b^2 \)
Now substituting \( a^2 = 3 \) back into the equation for \( b^2 \):

\[ b^2 = \frac{2}{3} \cdot 3 = 2 \]

### Step 4: Find the foci of the ellipse
The foci of the ellipse are located at \( (\pm ae, 0) \). First, we need to find \( e \):

\[ e = \frac{1}{\sqrt{3}} \implies ae = a \cdot \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{\sqrt{3}} = 1 \]

Thus, the foci are at \( (1, 0) \) and \( (-1, 0) \).

### Step 5: Circle centered at one of the foci
We choose the focus at \( (1, 0) \) for the circle, with a radius of \( \frac{2}{\sqrt{3}} \). The equation of the circle is:

\[ (x - 1)^2 + y^2 = \left(\frac{2}{\sqrt{3}}\right)^2 = \frac{4}{3} \]

### Step 6: Set up the equations for intersection
Now we have two equations:

1. Ellipse: \( \frac{x^2}{3} + \frac{y^2}{2} = 1 \)
2. Circle: \( (x - 1)^2 + y^2 = \frac{4}{3} \)

### Step 7: Solve the system of equations
From the circle's equation:

\[ (x - 1)^2 + y^2 = \frac{4}{3} \]

Expanding gives:

\[ x^2 - 2x + 1 + y^2 = \frac{4}{3} \]

Substituting \( y^2 = 2 - \frac{2x^2}{3} \) from the ellipse into the circle's equation:

\[ x^2 - 2x + 1 + \left(2 - \frac{2x^2}{3}\right) = \frac{4}{3} \]

Combining terms leads to a quadratic equation in \( x \).

### Step 8: Find the intersection points
Solving this quadratic will yield the \( x \) values of the intersection points. Once we find \( x_1 \) and \( x_2 \), we can find corresponding \( y_1 \) and \( y_2 \) using the ellipse equation.

### Step 9: Calculate the square of the distance between the points
Using the distance formula:

\[ d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2 \]

### Final Result
After calculating, we find that the square of the distance between the two intersection points is:

\[ \frac{16}{3} \]